${p_n}$ is such that $S_n = {1over p_1}+{1over p_2}+cdots+{1over p_n}$ converges. Prove $sigma_n =...
$begingroup$
Let ${p_n}$ denote a sequence such that:
$$
S_n = {1over p_1} + {1over p_2} + cdots + {1over p_n}
$$
converges.
Prove that:
$$
sigma_n = left(1+{1over p_1}right)left(1+{1over p_2}right)cdotsleft(1+{1over p_n}right)
$$
converges, where $n, p_n in Bbb N$.
Consider each bracket from $sigma_n$. By ${1over p_n} > 0$:
$$
forall k in Bbb N: left(1+{1over p_k}right) > 1
$$
So $sigma_n$ must be monotonically increasing by:
$$
{sigma_{n+1} over sigma_n} = left(1+{1over p_{n+1}}right) > 1
$$
To show a monotonic sequence is convergent it's sufficient to show that it's bounded above. Lets try to find the bound.
Recall:
$$
ln(1+x) le x iff (1+x) le e^x tag1
$$
So by $(1)$ we have:
$$
sigma_n le e^{S_n}
$$
But $S_n$ is convergent! And thus:
$$
sigma_n le e^L
$$
where:
$$
L = lim_{ntoinfty}S_n
$$
Which by monotone convergence theorem proves $sigma_n$ is convergent.
I'm kindly asking to verify my proof and point to the mistakes in case of any. Thank you!
calculus sequences-and-series limits proof-verification
asked Jan 18 at 13:58
romanroman
2,54721226
$endgroup$
|
show 2 more comments
$begingroup$
Let ${p_n}$ denote a sequence such that:
$$
S_n = {1over p_1} + {1over p_2} + cdots + {1over p_n}
$$
converges.
Prove that:
$$
sigma_n = left(1+{1over p_1}right)left(1+{1over p_2}right)cdotsleft(1+{1over p_n}right)
$$
converges, where $n, p_n in Bbb N$.
Consider each bracket from $sigma_n$. By ${1over p_n} > 0$:
$$
forall k in Bbb N: left(1+{1over p_k}right) > 1
$$
So $sigma_n$ must be monotonically increasing by:
$$
{sigma_{n+1} over sigma_n} = left(1+{1over p_{n+1}}right) > 1
$$
To show a monotonic sequence is convergent it's sufficient to show that it's bounded above. Lets try to find the bound.
Recall:
$$
ln(1+x) le x iff (1+x) le e^x tag1
$$
So by $(1)$ we have:
$$
sigma_n le e^{S_n}
$$
But $S_n$ is convergent! And thus:
$$
sigma_n le e^L
$$
where:
$$
L = lim_{ntoinfty}S_n
$$
Which by monotone convergence theorem proves $sigma_n$ is convergent.
I'm kindly asking to verify my proof and point to the mistakes in case of any. Thank you!
calculus sequences-and-series limits proof-verification
asked Jan 18 at 13:58
romanroman
2,54721226
$endgroup$
2
$begingroup$
If $p_n >0$, $sigma_n$ is easily proved to be increasing and the inequality $sigma_n leq e^{S_n} leq e^L$ is easy as well. Some passages are very strange (why do you need $1/p_n rightarrow 0$? What is your digression about the harmonic series?) but the crux seems quite correct.
$endgroup$
– Mindlack
Jan 18 at 14:06
1
$begingroup$
@Mindlack please notice a part in the question section saying $n, p_n in Bbb N$
$endgroup$
– roman
Jan 18 at 14:09
1
$begingroup$
@Mindlack as far as the other notices, I agree with you, I will update the post
$endgroup$
– roman
Jan 18 at 14:10
$begingroup$
Right, thank you. I edited.
$endgroup$
– Mindlack
Jan 18 at 14:11
3
$begingroup$
This is correct now. You can actually even write that each factor $(1+1/p_n)$ is lower than $e^{1/p_n}$, shortening the argument.
$endgroup$
– Mindlack
Jan 18 at 14:16
|
show 2 more comments
$begingroup$
Let ${p_n}$ denote a sequence such that:
$$
S_n = {1over p_1} + {1over p_2} + cdots + {1over p_n}
$$
converges.
Prove that:
$$
sigma_n = left(1+{1over p_1}right)left(1+{1over p_2}right)cdotsleft(1+{1over p_n}right)
$$
converges, where $n, p_n in Bbb N$.
Consider each bracket from $sigma_n$. By ${1over p_n} > 0$:
$$
forall k in Bbb N: left(1+{1over p_k}right) > 1
$$
So $sigma_n$ must be monotonically increasing by:
$$
{sigma_{n+1} over sigma_n} = left(1+{1over p_{n+1}}right) > 1
$$
To show a monotonic sequence is convergent it's sufficient to show that it's bounded above. Lets try to find the bound.
Recall:
$$
ln(1+x) le x iff (1+x) le e^x tag1
$$
So by $(1)$ we have:
$$
sigma_n le e^{S_n}
$$
But $S_n$ is convergent! And thus:
$$
sigma_n le e^L
$$
where:
$$
L = lim_{ntoinfty}S_n
$$
Which by monotone convergence theorem proves $sigma_n$ is convergent.
I'm kindly asking to verify my proof and point to the mistakes in case of any. Thank you!
calculus sequences-and-series limits proof-verification
asked Jan 18 at 13:58
romanroman
2,54721226
$endgroup$
Let ${p_n}$ denote a sequence such that:
$$
S_n = {1over p_1} + {1over p_2} + cdots + {1over p_n}
$$
converges.
Prove that:
$$
sigma_n = left(1+{1over p_1}right)left(1+{1over p_2}right)cdotsleft(1+{1over p_n}right)
$$
converges, where $n, p_n in Bbb N$.
Consider each bracket from $sigma_n$. By ${1over p_n} > 0$:
$$
forall k in Bbb N: left(1+{1over p_k}right) > 1
$$
So $sigma_n$ must be monotonically increasing by:
$$
{sigma_{n+1} over sigma_n} = left(1+{1over p_{n+1}}right) > 1
$$
To show a monotonic sequence is convergent it's sufficient to show that it's bounded above. Lets try to find the bound.
Recall:
$$
ln(1+x) le x iff (1+x) le e^x tag1
$$
So by $(1)$ we have:
$$
sigma_n le e^{S_n}
$$
But $S_n$ is convergent! And thus:
$$
sigma_n le e^L
$$
where:
$$
L = lim_{ntoinfty}S_n
$$
Which by monotone convergence theorem proves $sigma_n$ is convergent.
I'm kindly asking to verify my proof and point to the mistakes in case of any. Thank you!
calculus sequences-and-series limits proof-verification
calculus sequences-and-series limits proof-verification
asked Jan 18 at 13:58
romanroman
2,54721226
asked Jan 18 at 13:58
romanroman
2,54721226
edited Jan 18 at 14:29
roman
asked Jan 18 at 13:58
romanroman
2,54721226
asked Jan 18 at 13:58
romanroman
2,54721226
asked Jan 18 at 13:58
romanroman
2,54721226
2,54721226
2
$begingroup$
If $p_n >0$, $sigma_n$ is easily proved to be increasing and the inequality $sigma_n leq e^{S_n} leq e^L$ is easy as well. Some passages are very strange (why do you need $1/p_n rightarrow 0$? What is your digression about the harmonic series?) but the crux seems quite correct.
$endgroup$
– Mindlack
Jan 18 at 14:06
1
$begingroup$
@Mindlack please notice a part in the question section saying $n, p_n in Bbb N$
$endgroup$
– roman
Jan 18 at 14:09
1
$begingroup$
@Mindlack as far as the other notices, I agree with you, I will update the post
$endgroup$
– roman
Jan 18 at 14:10
$begingroup$
Right, thank you. I edited.
$endgroup$
– Mindlack
Jan 18 at 14:11
3
$begingroup$
This is correct now. You can actually even write that each factor $(1+1/p_n)$ is lower than $e^{1/p_n}$, shortening the argument.
$endgroup$
– Mindlack
Jan 18 at 14:16
|
show 2 more comments
2
$begingroup$
If $p_n >0$, $sigma_n$ is easily proved to be increasing and the inequality $sigma_n leq e^{S_n} leq e^L$ is easy as well. Some passages are very strange (why do you need $1/p_n rightarrow 0$? What is your digression about the harmonic series?) but the crux seems quite correct.
$endgroup$
– Mindlack
Jan 18 at 14:06
1
$begingroup$
@Mindlack please notice a part in the question section saying $n, p_n in Bbb N$
$endgroup$
– roman
Jan 18 at 14:09
1
$begingroup$
@Mindlack as far as the other notices, I agree with you, I will update the post
$endgroup$
– roman
Jan 18 at 14:10
$begingroup$
Right, thank you. I edited.
$endgroup$
– Mindlack
Jan 18 at 14:11
3
$begingroup$
This is correct now. You can actually even write that each factor $(1+1/p_n)$ is lower than $e^{1/p_n}$, shortening the argument.
$endgroup$
– Mindlack
Jan 18 at 14:16
2
2
$begingroup$
If $p_n >0$, $sigma_n$ is easily proved to be increasing and the inequality $sigma_n leq e^{S_n} leq e^L$ is easy as well. Some passages are very strange (why do you need $1/p_n rightarrow 0$? What is your digression about the harmonic series?) but the crux seems quite correct.
$endgroup$
– Mindlack
Jan 18 at 14:06
$begingroup$
If $p_n >0$, $sigma_n$ is easily proved to be increasing and the inequality $sigma_n leq e^{S_n} leq e^L$ is easy as well. Some passages are very strange (why do you need $1/p_n rightarrow 0$? What is your digression about the harmonic series?) but the crux seems quite correct.
$endgroup$
– Mindlack
Jan 18 at 14:06
1
1
$begingroup$
@Mindlack please notice a part in the question section saying $n, p_n in Bbb N$
$endgroup$
– roman
Jan 18 at 14:09
$begingroup$
@Mindlack please notice a part in the question section saying $n, p_n in Bbb N$
$endgroup$
– roman
Jan 18 at 14:09
1
1
$begingroup$
@Mindlack as far as the other notices, I agree with you, I will update the post
$endgroup$
– roman
Jan 18 at 14:10
$begingroup$
@Mindlack as far as the other notices, I agree with you, I will update the post
$endgroup$
– roman
Jan 18 at 14:10
$begingroup$
Right, thank you. I edited.
$endgroup$
– Mindlack
Jan 18 at 14:11
$begingroup$
Right, thank you. I edited.
$endgroup$
– Mindlack
Jan 18 at 14:11
3
3
$begingroup$
This is correct now. You can actually even write that each factor $(1+1/p_n)$ is lower than $e^{1/p_n}$, shortening the argument.
$endgroup$
– Mindlack
Jan 18 at 14:16
$begingroup$
This is correct now. You can actually even write that each factor $(1+1/p_n)$ is lower than $e^{1/p_n}$, shortening the argument.
$endgroup$
– Mindlack
Jan 18 at 14:16
|
show 2 more comments
0
active
oldest
votes
Your Answer
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%2f3078284%2fp-n-is-such-that-s-n-1-over-p-11-over-p-2-cdots1-over-p-n-conve%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
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%2f3078284%2fp-n-is-such-that-s-n-1-over-p-11-over-p-2-cdots1-over-p-n-conve%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
2
$begingroup$
If $p_n >0$, $sigma_n$ is easily proved to be increasing and the inequality $sigma_n leq e^{S_n} leq e^L$ is easy as well. Some passages are very strange (why do you need $1/p_n rightarrow 0$? What is your digression about the harmonic series?) but the crux seems quite correct.
$endgroup$
– Mindlack
Jan 18 at 14:06
1
$begingroup$
@Mindlack please notice a part in the question section saying $n, p_n in Bbb N$
$endgroup$
– roman
Jan 18 at 14:09
1
$begingroup$
@Mindlack as far as the other notices, I agree with you, I will update the post
$endgroup$
– roman
Jan 18 at 14:10
$begingroup$
Right, thank you. I edited.
$endgroup$
– Mindlack
Jan 18 at 14:11
3
$begingroup$
This is correct now. You can actually even write that each factor $(1+1/p_n)$ is lower than $e^{1/p_n}$, shortening the argument.
$endgroup$
– Mindlack
Jan 18 at 14:16