Let $sum_{k=1}^infty a_n$ be convergent show that $sum_{k=1}^infty n(a_n-a_{n+1})$ converges
Let $sumlimits_{k=1}^infty a_n$ be a convergent series where $a_ngeq0$ and $(a_n)$ is a monotone decreasing sequence prove that the series $sumlimits_{k=1}^infty n(a_n-a_{n+1})$ also converges.
What I tried :
Let $(A_n)$ be the sequence of partial sums of the series $sumlimits_{k=1}^infty a_n$ and $(B_n)$ be the sequence of partial sums of the series $sumlimits_{k=1}^infty n(a_n-a_{n+1})$.
Since $A_n=sumlimits_{k=1}^n a_k$ and $B_n=sumlimits_{k=1}^n k(a_k-a_{k+1})$ we get that:
begin{align}
B_n&=A_n-na_{n+1}\
&=(a_1-a_{n+1})+(a_2-a_{n+1})+...+(a_n-a_{n+1})\
&>(a_1-a_n)+(a_2-a_n)+...+(a_n-a_n)\
&=B_{n-1}
end{align}
we see that $(B_n)$ is a monotone increasing sequence $...(1)$
and
$B_n=A_n-na_{n+1}<A_n$ this implies that the sequence $(B_n)$ is bounded above ...(2)
Therefore (from (1) and (2)) the sequence $(B_n)$ converges so the series $sumlimits_{k=1}^infty n(a_n-a_{n+1})$ also converges
Is my proof correct?
real-analysis sequences-and-series proof-verification convergence
edited yesterday
Henning Makholm
237k16301536
asked yesterday
Maths Survivor
492219
|
show 3 more comments
Let $sumlimits_{k=1}^infty a_n$ be a convergent series where $a_ngeq0$ and $(a_n)$ is a monotone decreasing sequence prove that the series $sumlimits_{k=1}^infty n(a_n-a_{n+1})$ also converges.
What I tried :
Let $(A_n)$ be the sequence of partial sums of the series $sumlimits_{k=1}^infty a_n$ and $(B_n)$ be the sequence of partial sums of the series $sumlimits_{k=1}^infty n(a_n-a_{n+1})$.
Since $A_n=sumlimits_{k=1}^n a_k$ and $B_n=sumlimits_{k=1}^n k(a_k-a_{k+1})$ we get that:
begin{align}
B_n&=A_n-na_{n+1}\
&=(a_1-a_{n+1})+(a_2-a_{n+1})+...+(a_n-a_{n+1})\
&>(a_1-a_n)+(a_2-a_n)+...+(a_n-a_n)\
&=B_{n-1}
end{align}
we see that $(B_n)$ is a monotone increasing sequence $...(1)$
and
$B_n=A_n-na_{n+1}<A_n$ this implies that the sequence $(B_n)$ is bounded above ...(2)
Therefore (from (1) and (2)) the sequence $(B_n)$ converges so the series $sumlimits_{k=1}^infty n(a_n-a_{n+1})$ also converges
Is my proof correct?
real-analysis sequences-and-series proof-verification convergence
edited yesterday
Henning Makholm
237k16301536
asked yesterday
Maths Survivor
492219
Looks ok. Some simplifications: you know that $B_n - B_{n-1} = n(a_n - a_{n+1})$ which is $>0$. Another way to look at it: you know that $A_n$ converges and $B_n = A_n - na_{n+1}$. This means that the converge of $B_n$ is equivalent to showing $na_{n+1} to 0$ as $ntoinfty$.
– Winther
yesterday
Thank you , and from the second one do we get that the sum of $sum_{k=1}^infty n(a_n-a_{n+1})$ equals the sum of $sum_{k=1}^infty a_n$?
– Maths Survivor
yesterday
1
Taking the limit $ntoinfty$ of $B_n = A_n - na_{n+1}$. However disregard the last thing I said. The usual way of showing that $na_{n+1} to 0$ is to show that $sum n(a_n-a_{n+1})$ converges so that would be circular (see e.g. math.stackexchange.com/questions/383769/…). Your approach is good. btw that $B_n = A_n - na_{n+1}$ is a special case of summation by parts.
– Winther
yesterday
I can use another method to prove that $na_nto0$ when $n to infty$ without using the convergence of the series $sum_{k=1}^infty n(a_n-a_{n+1})$ , so from $B_n=A_n-na_n$ and since the sequences $(A_n)$ and $(na_n)$ converge when $n to infty $ implies that also the sequnce $(B_n)$ converges right?
– Maths Survivor
yesterday
Yes that's right
– Winther
yesterday
|
show 3 more comments
Let $sumlimits_{k=1}^infty a_n$ be a convergent series where $a_ngeq0$ and $(a_n)$ is a monotone decreasing sequence prove that the series $sumlimits_{k=1}^infty n(a_n-a_{n+1})$ also converges.
What I tried :
Let $(A_n)$ be the sequence of partial sums of the series $sumlimits_{k=1}^infty a_n$ and $(B_n)$ be the sequence of partial sums of the series $sumlimits_{k=1}^infty n(a_n-a_{n+1})$.
Since $A_n=sumlimits_{k=1}^n a_k$ and $B_n=sumlimits_{k=1}^n k(a_k-a_{k+1})$ we get that:
begin{align}
B_n&=A_n-na_{n+1}\
&=(a_1-a_{n+1})+(a_2-a_{n+1})+...+(a_n-a_{n+1})\
&>(a_1-a_n)+(a_2-a_n)+...+(a_n-a_n)\
&=B_{n-1}
end{align}
we see that $(B_n)$ is a monotone increasing sequence $...(1)$
and
$B_n=A_n-na_{n+1}<A_n$ this implies that the sequence $(B_n)$ is bounded above ...(2)
Therefore (from (1) and (2)) the sequence $(B_n)$ converges so the series $sumlimits_{k=1}^infty n(a_n-a_{n+1})$ also converges
Is my proof correct?
real-analysis sequences-and-series proof-verification convergence
edited yesterday
Henning Makholm
237k16301536
asked yesterday
Maths Survivor
492219
Let $sumlimits_{k=1}^infty a_n$ be a convergent series where $a_ngeq0$ and $(a_n)$ is a monotone decreasing sequence prove that the series $sumlimits_{k=1}^infty n(a_n-a_{n+1})$ also converges.
What I tried :
Let $(A_n)$ be the sequence of partial sums of the series $sumlimits_{k=1}^infty a_n$ and $(B_n)$ be the sequence of partial sums of the series $sumlimits_{k=1}^infty n(a_n-a_{n+1})$.
Since $A_n=sumlimits_{k=1}^n a_k$ and $B_n=sumlimits_{k=1}^n k(a_k-a_{k+1})$ we get that:
begin{align}
B_n&=A_n-na_{n+1}\
&=(a_1-a_{n+1})+(a_2-a_{n+1})+...+(a_n-a_{n+1})\
&>(a_1-a_n)+(a_2-a_n)+...+(a_n-a_n)\
&=B_{n-1}
end{align}
we see that $(B_n)$ is a monotone increasing sequence $...(1)$
and
$B_n=A_n-na_{n+1}<A_n$ this implies that the sequence $(B_n)$ is bounded above ...(2)
Therefore (from (1) and (2)) the sequence $(B_n)$ converges so the series $sumlimits_{k=1}^infty n(a_n-a_{n+1})$ also converges
Is my proof correct?
real-analysis sequences-and-series proof-verification convergence
real-analysis sequences-and-series proof-verification convergence
edited yesterday
Henning Makholm
237k16301536
asked yesterday
Maths Survivor
492219
edited yesterday
Henning Makholm
237k16301536
asked yesterday
Maths Survivor
492219
edited yesterday
Henning Makholm
237k16301536
edited yesterday
Henning Makholm
237k16301536
edited yesterday
Henning Makholm
237k16301536
237k16301536
asked yesterday
Maths Survivor
492219
asked yesterday
Maths Survivor
492219
asked yesterday
Maths Survivor
492219
492219
Looks ok. Some simplifications: you know that $B_n - B_{n-1} = n(a_n - a_{n+1})$ which is $>0$. Another way to look at it: you know that $A_n$ converges and $B_n = A_n - na_{n+1}$. This means that the converge of $B_n$ is equivalent to showing $na_{n+1} to 0$ as $ntoinfty$.
– Winther
yesterday
Thank you , and from the second one do we get that the sum of $sum_{k=1}^infty n(a_n-a_{n+1})$ equals the sum of $sum_{k=1}^infty a_n$?
– Maths Survivor
yesterday
1
Taking the limit $ntoinfty$ of $B_n = A_n - na_{n+1}$. However disregard the last thing I said. The usual way of showing that $na_{n+1} to 0$ is to show that $sum n(a_n-a_{n+1})$ converges so that would be circular (see e.g. math.stackexchange.com/questions/383769/…). Your approach is good. btw that $B_n = A_n - na_{n+1}$ is a special case of summation by parts.
– Winther
yesterday
I can use another method to prove that $na_nto0$ when $n to infty$ without using the convergence of the series $sum_{k=1}^infty n(a_n-a_{n+1})$ , so from $B_n=A_n-na_n$ and since the sequences $(A_n)$ and $(na_n)$ converge when $n to infty $ implies that also the sequnce $(B_n)$ converges right?
– Maths Survivor
yesterday
Yes that's right
– Winther
yesterday
|
show 3 more comments
Looks ok. Some simplifications: you know that $B_n - B_{n-1} = n(a_n - a_{n+1})$ which is $>0$. Another way to look at it: you know that $A_n$ converges and $B_n = A_n - na_{n+1}$. This means that the converge of $B_n$ is equivalent to showing $na_{n+1} to 0$ as $ntoinfty$.
– Winther
yesterday
Thank you , and from the second one do we get that the sum of $sum_{k=1}^infty n(a_n-a_{n+1})$ equals the sum of $sum_{k=1}^infty a_n$?
– Maths Survivor
yesterday
1
Taking the limit $ntoinfty$ of $B_n = A_n - na_{n+1}$. However disregard the last thing I said. The usual way of showing that $na_{n+1} to 0$ is to show that $sum n(a_n-a_{n+1})$ converges so that would be circular (see e.g. math.stackexchange.com/questions/383769/…). Your approach is good. btw that $B_n = A_n - na_{n+1}$ is a special case of summation by parts.
– Winther
yesterday
I can use another method to prove that $na_nto0$ when $n to infty$ without using the convergence of the series $sum_{k=1}^infty n(a_n-a_{n+1})$ , so from $B_n=A_n-na_n$ and since the sequences $(A_n)$ and $(na_n)$ converge when $n to infty $ implies that also the sequnce $(B_n)$ converges right?
– Maths Survivor
yesterday
Yes that's right
– Winther
yesterday
Looks ok. Some simplifications: you know that $B_n - B_{n-1} = n(a_n - a_{n+1})$ which is $>0$. Another way to look at it: you know that $A_n$ converges and $B_n = A_n - na_{n+1}$. This means that the converge of $B_n$ is equivalent to showing $na_{n+1} to 0$ as $ntoinfty$.
– Winther
yesterday
Looks ok. Some simplifications: you know that $B_n - B_{n-1} = n(a_n - a_{n+1})$ which is $>0$. Another way to look at it: you know that $A_n$ converges and $B_n = A_n - na_{n+1}$. This means that the converge of $B_n$ is equivalent to showing $na_{n+1} to 0$ as $ntoinfty$.
– Winther
yesterday
Thank you , and from the second one do we get that the sum of $sum_{k=1}^infty n(a_n-a_{n+1})$ equals the sum of $sum_{k=1}^infty a_n$?
– Maths Survivor
yesterday
Thank you , and from the second one do we get that the sum of $sum_{k=1}^infty n(a_n-a_{n+1})$ equals the sum of $sum_{k=1}^infty a_n$?
– Maths Survivor
yesterday
1
1
Taking the limit $ntoinfty$ of $B_n = A_n - na_{n+1}$. However disregard the last thing I said. The usual way of showing that $na_{n+1} to 0$ is to show that $sum n(a_n-a_{n+1})$ converges so that would be circular (see e.g. math.stackexchange.com/questions/383769/…). Your approach is good. btw that $B_n = A_n - na_{n+1}$ is a special case of summation by parts.
– Winther
yesterday
Taking the limit $ntoinfty$ of $B_n = A_n - na_{n+1}$. However disregard the last thing I said. The usual way of showing that $na_{n+1} to 0$ is to show that $sum n(a_n-a_{n+1})$ converges so that would be circular (see e.g. math.stackexchange.com/questions/383769/…). Your approach is good. btw that $B_n = A_n - na_{n+1}$ is a special case of summation by parts.
– Winther
yesterday
I can use another method to prove that $na_nto0$ when $n to infty$ without using the convergence of the series $sum_{k=1}^infty n(a_n-a_{n+1})$ , so from $B_n=A_n-na_n$ and since the sequences $(A_n)$ and $(na_n)$ converge when $n to infty $ implies that also the sequnce $(B_n)$ converges right?
– Maths Survivor
yesterday
I can use another method to prove that $na_nto0$ when $n to infty$ without using the convergence of the series $sum_{k=1}^infty n(a_n-a_{n+1})$ , so from $B_n=A_n-na_n$ and since the sequences $(A_n)$ and $(na_n)$ converge when $n to infty $ implies that also the sequnce $(B_n)$ converges right?
– Maths Survivor
yesterday
Yes that's right
– Winther
yesterday
Yes that's right
– Winther
yesterday
|
show 3 more comments
2 Answers
2
active
oldest
votes
Some comments:
The identity $B_n = A_n - n a_{n+1}$ is not quite obvious enough to state without proof. A quick induction proof would work, as would writing out the sums using ellipses (i.e. the symbol "$ldots$") and simplifying.
Showing $B_n < A_n$ establishes an upper bound based on $n$, which is not allowed (e.g. $B_n le B_n$ always!). As $A_n$ (being the partial sums of a convergent series) is convergent, you can easily establish an upper bound (especially when you consider the fact that $A_n$ is increasing).
You can more quickly establish that $B_n$ is increasing by observing that it is the sum of positive numbers.
You could also further note that, if $lim B_n < lim A_n$, then $a_n$ is approximately a multiple of the harmonic series, which is divergent, thus the two series share the same sum. (This is not a criticism, just something worth noting).
answered yesterday
Theo Bendit
16.4k12148
1
"then $a_n$ is approximately a multiple of the harmonic series" what does this mean? Can you give more explanation to this sentence?
– Maths Survivor
yesterday
1
@MathsSurvivor If $A_n to A$ and $B_n to B$ with $A neq B$, then $$frac{a_{n+1}}{frac{1}{n}} = na_{n+1} = A_n - B_n to A - B neq 0,$$ so by the limit comparison test, $a_{n+1}$, when summed, is a divergent sequence. This contradicts $A_n$ converging.
– Theo Bendit
yesterday
add a comment |
Note that we can use $n=sum_{k=1}^n (1)$ to write
$$begin{align}
sum_{n=1}^N n(a_{n+1}-a_n)&=sum_{n=1}^N sum_{k=1}^n(a_{n+1}-a_n)\\
&=sum_{k=1}^N sum_{n=k}^N (a_{n+1}-a_n)\\
&=sum_{k=1}^N (a_{N+1}-a_k)\\
&=Na_{N+1}-sum_{k=1}^N a_ktag1
end{align}$$
Inasmuch as $a_nge 0$ monotonically decreases to $0$, and $sum_{k=1}^n a_n<infty$, we have $lim_{ntoinfty }na_n=0$. Hence, using $(1)$, we see that
$$begin{align}
lim_{Nto infty }sum_{n=1}^N n(a_{n+1}-a_n)&=lim_{Ntoinfty}left(Na_{N+1}-sum_{k=1}^N a_kright)\\
&=-sum_{n=1}^infty a_n
end{align}$$
from which we conclude that $sum_{n=1}^infty n(a_{n+1}-a_n)$ converges.
answered yesterday
Mark Viola
130k1273170
Winther's comment on the question suggests that showing $lim n a_n = 0$ is often done by first establishing $sum n(a_n - a_{n+1}) = sum a_n$. Out of curiosity, how would you establish $lim n a_n = 0$?
– Theo Bendit
yesterday
Note that $$(2n)a_{2n}le 2sum_{n+1}^{2n}a_kto 0$$since $a_nge0$ is monotonic and $sum_n a_n<infty.$.
– Mark Viola
yesterday
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052073%2flet-sum-k-1-infty-a-n-be-convergent-show-that-sum-k-1-infty-na-n-a%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Some comments:
The identity $B_n = A_n - n a_{n+1}$ is not quite obvious enough to state without proof. A quick induction proof would work, as would writing out the sums using ellipses (i.e. the symbol "$ldots$") and simplifying.
Showing $B_n < A_n$ establishes an upper bound based on $n$, which is not allowed (e.g. $B_n le B_n$ always!). As $A_n$ (being the partial sums of a convergent series) is convergent, you can easily establish an upper bound (especially when you consider the fact that $A_n$ is increasing).
You can more quickly establish that $B_n$ is increasing by observing that it is the sum of positive numbers.
You could also further note that, if $lim B_n < lim A_n$, then $a_n$ is approximately a multiple of the harmonic series, which is divergent, thus the two series share the same sum. (This is not a criticism, just something worth noting).
answered yesterday
Theo Bendit
16.4k12148
1
"then $a_n$ is approximately a multiple of the harmonic series" what does this mean? Can you give more explanation to this sentence?
– Maths Survivor
yesterday
1
@MathsSurvivor If $A_n to A$ and $B_n to B$ with $A neq B$, then $$frac{a_{n+1}}{frac{1}{n}} = na_{n+1} = A_n - B_n to A - B neq 0,$$ so by the limit comparison test, $a_{n+1}$, when summed, is a divergent sequence. This contradicts $A_n$ converging.
– Theo Bendit
yesterday
add a comment |
Some comments:
The identity $B_n = A_n - n a_{n+1}$ is not quite obvious enough to state without proof. A quick induction proof would work, as would writing out the sums using ellipses (i.e. the symbol "$ldots$") and simplifying.
Showing $B_n < A_n$ establishes an upper bound based on $n$, which is not allowed (e.g. $B_n le B_n$ always!). As $A_n$ (being the partial sums of a convergent series) is convergent, you can easily establish an upper bound (especially when you consider the fact that $A_n$ is increasing).
You can more quickly establish that $B_n$ is increasing by observing that it is the sum of positive numbers.
You could also further note that, if $lim B_n < lim A_n$, then $a_n$ is approximately a multiple of the harmonic series, which is divergent, thus the two series share the same sum. (This is not a criticism, just something worth noting).
answered yesterday
Theo Bendit
16.4k12148
1
"then $a_n$ is approximately a multiple of the harmonic series" what does this mean? Can you give more explanation to this sentence?
– Maths Survivor
yesterday
1
@MathsSurvivor If $A_n to A$ and $B_n to B$ with $A neq B$, then $$frac{a_{n+1}}{frac{1}{n}} = na_{n+1} = A_n - B_n to A - B neq 0,$$ so by the limit comparison test, $a_{n+1}$, when summed, is a divergent sequence. This contradicts $A_n$ converging.
– Theo Bendit
yesterday
add a comment |
Some comments:
The identity $B_n = A_n - n a_{n+1}$ is not quite obvious enough to state without proof. A quick induction proof would work, as would writing out the sums using ellipses (i.e. the symbol "$ldots$") and simplifying.
Showing $B_n < A_n$ establishes an upper bound based on $n$, which is not allowed (e.g. $B_n le B_n$ always!). As $A_n$ (being the partial sums of a convergent series) is convergent, you can easily establish an upper bound (especially when you consider the fact that $A_n$ is increasing).
You can more quickly establish that $B_n$ is increasing by observing that it is the sum of positive numbers.
You could also further note that, if $lim B_n < lim A_n$, then $a_n$ is approximately a multiple of the harmonic series, which is divergent, thus the two series share the same sum. (This is not a criticism, just something worth noting).
answered yesterday
Theo Bendit
16.4k12148
Some comments:
The identity $B_n = A_n - n a_{n+1}$ is not quite obvious enough to state without proof. A quick induction proof would work, as would writing out the sums using ellipses (i.e. the symbol "$ldots$") and simplifying.
Showing $B_n < A_n$ establishes an upper bound based on $n$, which is not allowed (e.g. $B_n le B_n$ always!). As $A_n$ (being the partial sums of a convergent series) is convergent, you can easily establish an upper bound (especially when you consider the fact that $A_n$ is increasing).
You can more quickly establish that $B_n$ is increasing by observing that it is the sum of positive numbers.
You could also further note that, if $lim B_n < lim A_n$, then $a_n$ is approximately a multiple of the harmonic series, which is divergent, thus the two series share the same sum. (This is not a criticism, just something worth noting).
answered yesterday
Theo Bendit
16.4k12148
answered yesterday
Theo Bendit
16.4k12148
answered yesterday
Theo Bendit
16.4k12148
answered yesterday
Theo Bendit
16.4k12148
16.4k12148
1
"then $a_n$ is approximately a multiple of the harmonic series" what does this mean? Can you give more explanation to this sentence?
– Maths Survivor
yesterday
1
@MathsSurvivor If $A_n to A$ and $B_n to B$ with $A neq B$, then $$frac{a_{n+1}}{frac{1}{n}} = na_{n+1} = A_n - B_n to A - B neq 0,$$ so by the limit comparison test, $a_{n+1}$, when summed, is a divergent sequence. This contradicts $A_n$ converging.
– Theo Bendit
yesterday
add a comment |
1
"then $a_n$ is approximately a multiple of the harmonic series" what does this mean? Can you give more explanation to this sentence?
– Maths Survivor
yesterday
1
@MathsSurvivor If $A_n to A$ and $B_n to B$ with $A neq B$, then $$frac{a_{n+1}}{frac{1}{n}} = na_{n+1} = A_n - B_n to A - B neq 0,$$ so by the limit comparison test, $a_{n+1}$, when summed, is a divergent sequence. This contradicts $A_n$ converging.
– Theo Bendit
yesterday
1
1
"then $a_n$ is approximately a multiple of the harmonic series" what does this mean? Can you give more explanation to this sentence?
– Maths Survivor
yesterday
"then $a_n$ is approximately a multiple of the harmonic series" what does this mean? Can you give more explanation to this sentence?
– Maths Survivor
yesterday
1
1
@MathsSurvivor If $A_n to A$ and $B_n to B$ with $A neq B$, then $$frac{a_{n+1}}{frac{1}{n}} = na_{n+1} = A_n - B_n to A - B neq 0,$$ so by the limit comparison test, $a_{n+1}$, when summed, is a divergent sequence. This contradicts $A_n$ converging.
– Theo Bendit
yesterday
@MathsSurvivor If $A_n to A$ and $B_n to B$ with $A neq B$, then $$frac{a_{n+1}}{frac{1}{n}} = na_{n+1} = A_n - B_n to A - B neq 0,$$ so by the limit comparison test, $a_{n+1}$, when summed, is a divergent sequence. This contradicts $A_n$ converging.
– Theo Bendit
yesterday
add a comment |
Note that we can use $n=sum_{k=1}^n (1)$ to write
$$begin{align}
sum_{n=1}^N n(a_{n+1}-a_n)&=sum_{n=1}^N sum_{k=1}^n(a_{n+1}-a_n)\\
&=sum_{k=1}^N sum_{n=k}^N (a_{n+1}-a_n)\\
&=sum_{k=1}^N (a_{N+1}-a_k)\\
&=Na_{N+1}-sum_{k=1}^N a_ktag1
end{align}$$
Inasmuch as $a_nge 0$ monotonically decreases to $0$, and $sum_{k=1}^n a_n<infty$, we have $lim_{ntoinfty }na_n=0$. Hence, using $(1)$, we see that
$$begin{align}
lim_{Nto infty }sum_{n=1}^N n(a_{n+1}-a_n)&=lim_{Ntoinfty}left(Na_{N+1}-sum_{k=1}^N a_kright)\\
&=-sum_{n=1}^infty a_n
end{align}$$
from which we conclude that $sum_{n=1}^infty n(a_{n+1}-a_n)$ converges.
answered yesterday
Mark Viola
130k1273170
Winther's comment on the question suggests that showing $lim n a_n = 0$ is often done by first establishing $sum n(a_n - a_{n+1}) = sum a_n$. Out of curiosity, how would you establish $lim n a_n = 0$?
– Theo Bendit
yesterday
Note that $$(2n)a_{2n}le 2sum_{n+1}^{2n}a_kto 0$$since $a_nge0$ is monotonic and $sum_n a_n<infty.$.
– Mark Viola
yesterday
add a comment |
Note that we can use $n=sum_{k=1}^n (1)$ to write
$$begin{align}
sum_{n=1}^N n(a_{n+1}-a_n)&=sum_{n=1}^N sum_{k=1}^n(a_{n+1}-a_n)\\
&=sum_{k=1}^N sum_{n=k}^N (a_{n+1}-a_n)\\
&=sum_{k=1}^N (a_{N+1}-a_k)\\
&=Na_{N+1}-sum_{k=1}^N a_ktag1
end{align}$$
Inasmuch as $a_nge 0$ monotonically decreases to $0$, and $sum_{k=1}^n a_n<infty$, we have $lim_{ntoinfty }na_n=0$. Hence, using $(1)$, we see that
$$begin{align}
lim_{Nto infty }sum_{n=1}^N n(a_{n+1}-a_n)&=lim_{Ntoinfty}left(Na_{N+1}-sum_{k=1}^N a_kright)\\
&=-sum_{n=1}^infty a_n
end{align}$$
from which we conclude that $sum_{n=1}^infty n(a_{n+1}-a_n)$ converges.
answered yesterday
Mark Viola
130k1273170
Winther's comment on the question suggests that showing $lim n a_n = 0$ is often done by first establishing $sum n(a_n - a_{n+1}) = sum a_n$. Out of curiosity, how would you establish $lim n a_n = 0$?
– Theo Bendit
yesterday
Note that $$(2n)a_{2n}le 2sum_{n+1}^{2n}a_kto 0$$since $a_nge0$ is monotonic and $sum_n a_n<infty.$.
– Mark Viola
yesterday
add a comment |
Note that we can use $n=sum_{k=1}^n (1)$ to write
$$begin{align}
sum_{n=1}^N n(a_{n+1}-a_n)&=sum_{n=1}^N sum_{k=1}^n(a_{n+1}-a_n)\\
&=sum_{k=1}^N sum_{n=k}^N (a_{n+1}-a_n)\\
&=sum_{k=1}^N (a_{N+1}-a_k)\\
&=Na_{N+1}-sum_{k=1}^N a_ktag1
end{align}$$
Inasmuch as $a_nge 0$ monotonically decreases to $0$, and $sum_{k=1}^n a_n<infty$, we have $lim_{ntoinfty }na_n=0$. Hence, using $(1)$, we see that
$$begin{align}
lim_{Nto infty }sum_{n=1}^N n(a_{n+1}-a_n)&=lim_{Ntoinfty}left(Na_{N+1}-sum_{k=1}^N a_kright)\\
&=-sum_{n=1}^infty a_n
end{align}$$
from which we conclude that $sum_{n=1}^infty n(a_{n+1}-a_n)$ converges.
answered yesterday
Mark Viola
130k1273170
Note that we can use $n=sum_{k=1}^n (1)$ to write
$$begin{align}
sum_{n=1}^N n(a_{n+1}-a_n)&=sum_{n=1}^N sum_{k=1}^n(a_{n+1}-a_n)\\
&=sum_{k=1}^N sum_{n=k}^N (a_{n+1}-a_n)\\
&=sum_{k=1}^N (a_{N+1}-a_k)\\
&=Na_{N+1}-sum_{k=1}^N a_ktag1
end{align}$$
Inasmuch as $a_nge 0$ monotonically decreases to $0$, and $sum_{k=1}^n a_n<infty$, we have $lim_{ntoinfty }na_n=0$. Hence, using $(1)$, we see that
$$begin{align}
lim_{Nto infty }sum_{n=1}^N n(a_{n+1}-a_n)&=lim_{Ntoinfty}left(Na_{N+1}-sum_{k=1}^N a_kright)\\
&=-sum_{n=1}^infty a_n
end{align}$$
from which we conclude that $sum_{n=1}^infty n(a_{n+1}-a_n)$ converges.
answered yesterday
Mark Viola
130k1273170
answered yesterday
Mark Viola
130k1273170
answered yesterday
Mark Viola
130k1273170
answered yesterday
Mark Viola
130k1273170
130k1273170
Winther's comment on the question suggests that showing $lim n a_n = 0$ is often done by first establishing $sum n(a_n - a_{n+1}) = sum a_n$. Out of curiosity, how would you establish $lim n a_n = 0$?
– Theo Bendit
yesterday
Note that $$(2n)a_{2n}le 2sum_{n+1}^{2n}a_kto 0$$since $a_nge0$ is monotonic and $sum_n a_n<infty.$.
– Mark Viola
yesterday
add a comment |
Winther's comment on the question suggests that showing $lim n a_n = 0$ is often done by first establishing $sum n(a_n - a_{n+1}) = sum a_n$. Out of curiosity, how would you establish $lim n a_n = 0$?
– Theo Bendit
yesterday
Note that $$(2n)a_{2n}le 2sum_{n+1}^{2n}a_kto 0$$since $a_nge0$ is monotonic and $sum_n a_n<infty.$.
– Mark Viola
yesterday
Winther's comment on the question suggests that showing $lim n a_n = 0$ is often done by first establishing $sum n(a_n - a_{n+1}) = sum a_n$. Out of curiosity, how would you establish $lim n a_n = 0$?
– Theo Bendit
yesterday
Winther's comment on the question suggests that showing $lim n a_n = 0$ is often done by first establishing $sum n(a_n - a_{n+1}) = sum a_n$. Out of curiosity, how would you establish $lim n a_n = 0$?
– Theo Bendit
yesterday
Note that $$(2n)a_{2n}le 2sum_{n+1}^{2n}a_kto 0$$since $a_nge0$ is monotonic and $sum_n a_n<infty.$.
– Mark Viola
yesterday
Note that $$(2n)a_{2n}le 2sum_{n+1}^{2n}a_kto 0$$since $a_nge0$ is monotonic and $sum_n a_n<infty.$.
– Mark Viola
yesterday
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3052073%2flet-sum-k-1-infty-a-n-be-convergent-show-that-sum-k-1-infty-na-n-a%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Looks ok. Some simplifications: you know that $B_n - B_{n-1} = n(a_n - a_{n+1})$ which is $>0$. Another way to look at it: you know that $A_n$ converges and $B_n = A_n - na_{n+1}$. This means that the converge of $B_n$ is equivalent to showing $na_{n+1} to 0$ as $ntoinfty$.
– Winther
yesterday
Thank you , and from the second one do we get that the sum of $sum_{k=1}^infty n(a_n-a_{n+1})$ equals the sum of $sum_{k=1}^infty a_n$?
– Maths Survivor
yesterday
1
Taking the limit $ntoinfty$ of $B_n = A_n - na_{n+1}$. However disregard the last thing I said. The usual way of showing that $na_{n+1} to 0$ is to show that $sum n(a_n-a_{n+1})$ converges so that would be circular (see e.g. math.stackexchange.com/questions/383769/…). Your approach is good. btw that $B_n = A_n - na_{n+1}$ is a special case of summation by parts.
– Winther
yesterday
I can use another method to prove that $na_nto0$ when $n to infty$ without using the convergence of the series $sum_{k=1}^infty n(a_n-a_{n+1})$ , so from $B_n=A_n-na_n$ and since the sequences $(A_n)$ and $(na_n)$ converge when $n to infty $ implies that also the sequnce $(B_n)$ converges right?
– Maths Survivor
yesterday
Yes that's right
– Winther
yesterday