Divergence for Series with $|a_n| ge b_n ge 0$
$begingroup$
first of all i hope you get my point since English is not my native.
I know the series $$ y = sum_{k=1}^infty frac{1}{k} $$ is divergent.
Now I want to check if the series $$x=sum_{k=1}^infty frac{1}{2(k+1)} $$ is divergent too.
I "have" to use the criteria with $$ |a_k| ge d_k ge 0 $$ in Germany it is called Minorante.
So when $sum d_k$ is divergent obviously $sum a_k$ is divergent too.
In my case with an indexshift we can put $$ sum_{k=1}^infty frac{1}{2(k+1)} $$ into $$ 1/2 * sum_{k=2}^infty 1/k $$
We see when we shift the Index in the other formula $$ y = sum_{k=1}^infty 1/k $$ to $$ y = sum_{k=2}^infty 1/k $$ we have nearly the same expression only that we have a *1/2 on $x_k$.
My question is, the definition is $$ |a_k| ge d_k ge 0. $$ For our case $$ x ge y ge 0 $$
Why is our $x_k = a_k $ and our $y_k = d_k $?
Because $ y = sum_{k=2}^infty 1/k $ is $gt$ than $ x = 1/2*sum_{k=2}^infty 1/k $
Or do we ignore the 1/2*? Then it would be at least equal and the definition would be okay.
My solution from the university says: We know $y = sum_{k=2}^infty 1/k$ is divergent so we can conclude with the defintion $|a_k| ge d_k ge 0$ that $x = sum_{k=1}^infty 1/(2(k+1))$ is divergent too.
sequences-and-series convergence divergent-series
edited Mar 16 at 18:42
Robert Howard
2,2933935
asked Jan 16 at 14:53
BrainOverflowBrainOverflow
33
$endgroup$
add a comment |
$begingroup$
first of all i hope you get my point since English is not my native.
I know the series $$ y = sum_{k=1}^infty frac{1}{k} $$ is divergent.
Now I want to check if the series $$x=sum_{k=1}^infty frac{1}{2(k+1)} $$ is divergent too.
I "have" to use the criteria with $$ |a_k| ge d_k ge 0 $$ in Germany it is called Minorante.
So when $sum d_k$ is divergent obviously $sum a_k$ is divergent too.
In my case with an indexshift we can put $$ sum_{k=1}^infty frac{1}{2(k+1)} $$ into $$ 1/2 * sum_{k=2}^infty 1/k $$
We see when we shift the Index in the other formula $$ y = sum_{k=1}^infty 1/k $$ to $$ y = sum_{k=2}^infty 1/k $$ we have nearly the same expression only that we have a *1/2 on $x_k$.
My question is, the definition is $$ |a_k| ge d_k ge 0. $$ For our case $$ x ge y ge 0 $$
Why is our $x_k = a_k $ and our $y_k = d_k $?
Because $ y = sum_{k=2}^infty 1/k $ is $gt$ than $ x = 1/2*sum_{k=2}^infty 1/k $
Or do we ignore the 1/2*? Then it would be at least equal and the definition would be okay.
My solution from the university says: We know $y = sum_{k=2}^infty 1/k$ is divergent so we can conclude with the defintion $|a_k| ge d_k ge 0$ that $x = sum_{k=1}^infty 1/(2(k+1))$ is divergent too.
sequences-and-series convergence divergent-series
edited Mar 16 at 18:42
Robert Howard
2,2933935
asked Jan 16 at 14:53
BrainOverflowBrainOverflow
33
$endgroup$
1
$begingroup$
Hello and welcome to math.stackexchange. There are several issues with your question. (1) If $|a_k| ge d_k > 0$ and the series $sum_k d_k$ diverges, it does not follow that the series $sum_k a_k$ diverges. This is only true if all (or almost all) $a_k$ are positive or negative. (2) You can't write something like $y_k = sum_k a_k$, since the index $k$ is a dummy variable and the result of the series does not depend on $k$. To understand the argument in your notes/solution key, write down what it would mean for the first 4 or 6 terms.
$endgroup$
– Hans Engler
Jan 16 at 15:05
$begingroup$
@HansEngler to (1) Since we only use natural Numbers I think we ignore this since it will always be positive right? (2) Yeah sorry I know that. I don't know why I wrote y_k etc. I corrected it. Well, for y we would have 1/2 + 1/3 + 1/4 ... For x we would have 1/4+ 1/6 + 1/8...
$endgroup$
– BrainOverflow
Jan 16 at 15:14
add a comment |
$begingroup$
first of all i hope you get my point since English is not my native.
I know the series $$ y = sum_{k=1}^infty frac{1}{k} $$ is divergent.
Now I want to check if the series $$x=sum_{k=1}^infty frac{1}{2(k+1)} $$ is divergent too.
I "have" to use the criteria with $$ |a_k| ge d_k ge 0 $$ in Germany it is called Minorante.
So when $sum d_k$ is divergent obviously $sum a_k$ is divergent too.
In my case with an indexshift we can put $$ sum_{k=1}^infty frac{1}{2(k+1)} $$ into $$ 1/2 * sum_{k=2}^infty 1/k $$
We see when we shift the Index in the other formula $$ y = sum_{k=1}^infty 1/k $$ to $$ y = sum_{k=2}^infty 1/k $$ we have nearly the same expression only that we have a *1/2 on $x_k$.
My question is, the definition is $$ |a_k| ge d_k ge 0. $$ For our case $$ x ge y ge 0 $$
Why is our $x_k = a_k $ and our $y_k = d_k $?
Because $ y = sum_{k=2}^infty 1/k $ is $gt$ than $ x = 1/2*sum_{k=2}^infty 1/k $
Or do we ignore the 1/2*? Then it would be at least equal and the definition would be okay.
My solution from the university says: We know $y = sum_{k=2}^infty 1/k$ is divergent so we can conclude with the defintion $|a_k| ge d_k ge 0$ that $x = sum_{k=1}^infty 1/(2(k+1))$ is divergent too.
sequences-and-series convergence divergent-series
edited Mar 16 at 18:42
Robert Howard
2,2933935
asked Jan 16 at 14:53
BrainOverflowBrainOverflow
33
$endgroup$
first of all i hope you get my point since English is not my native.
I know the series $$ y = sum_{k=1}^infty frac{1}{k} $$ is divergent.
Now I want to check if the series $$x=sum_{k=1}^infty frac{1}{2(k+1)} $$ is divergent too.
I "have" to use the criteria with $$ |a_k| ge d_k ge 0 $$ in Germany it is called Minorante.
So when $sum d_k$ is divergent obviously $sum a_k$ is divergent too.
In my case with an indexshift we can put $$ sum_{k=1}^infty frac{1}{2(k+1)} $$ into $$ 1/2 * sum_{k=2}^infty 1/k $$
We see when we shift the Index in the other formula $$ y = sum_{k=1}^infty 1/k $$ to $$ y = sum_{k=2}^infty 1/k $$ we have nearly the same expression only that we have a *1/2 on $x_k$.
My question is, the definition is $$ |a_k| ge d_k ge 0. $$ For our case $$ x ge y ge 0 $$
Why is our $x_k = a_k $ and our $y_k = d_k $?
Because $ y = sum_{k=2}^infty 1/k $ is $gt$ than $ x = 1/2*sum_{k=2}^infty 1/k $
Or do we ignore the 1/2*? Then it would be at least equal and the definition would be okay.
My solution from the university says: We know $y = sum_{k=2}^infty 1/k$ is divergent so we can conclude with the defintion $|a_k| ge d_k ge 0$ that $x = sum_{k=1}^infty 1/(2(k+1))$ is divergent too.
sequences-and-series convergence divergent-series
sequences-and-series convergence divergent-series
edited Mar 16 at 18:42
Robert Howard
2,2933935
asked Jan 16 at 14:53
BrainOverflowBrainOverflow
33
edited Mar 16 at 18:42
Robert Howard
2,2933935
asked Jan 16 at 14:53
BrainOverflowBrainOverflow
33
edited Mar 16 at 18:42
Robert Howard
2,2933935
edited Mar 16 at 18:42
Robert Howard
2,2933935
edited Mar 16 at 18:42
Robert Howard
2,2933935
2,2933935
asked Jan 16 at 14:53
BrainOverflowBrainOverflow
33
asked Jan 16 at 14:53
BrainOverflowBrainOverflow
33
asked Jan 16 at 14:53
BrainOverflowBrainOverflow
33
33
1
$begingroup$
Hello and welcome to math.stackexchange. There are several issues with your question. (1) If $|a_k| ge d_k > 0$ and the series $sum_k d_k$ diverges, it does not follow that the series $sum_k a_k$ diverges. This is only true if all (or almost all) $a_k$ are positive or negative. (2) You can't write something like $y_k = sum_k a_k$, since the index $k$ is a dummy variable and the result of the series does not depend on $k$. To understand the argument in your notes/solution key, write down what it would mean for the first 4 or 6 terms.
$endgroup$
– Hans Engler
Jan 16 at 15:05
$begingroup$
@HansEngler to (1) Since we only use natural Numbers I think we ignore this since it will always be positive right? (2) Yeah sorry I know that. I don't know why I wrote y_k etc. I corrected it. Well, for y we would have 1/2 + 1/3 + 1/4 ... For x we would have 1/4+ 1/6 + 1/8...
$endgroup$
– BrainOverflow
Jan 16 at 15:14
add a comment |
1
$begingroup$
Hello and welcome to math.stackexchange. There are several issues with your question. (1) If $|a_k| ge d_k > 0$ and the series $sum_k d_k$ diverges, it does not follow that the series $sum_k a_k$ diverges. This is only true if all (or almost all) $a_k$ are positive or negative. (2) You can't write something like $y_k = sum_k a_k$, since the index $k$ is a dummy variable and the result of the series does not depend on $k$. To understand the argument in your notes/solution key, write down what it would mean for the first 4 or 6 terms.
$endgroup$
– Hans Engler
Jan 16 at 15:05
$begingroup$
@HansEngler to (1) Since we only use natural Numbers I think we ignore this since it will always be positive right? (2) Yeah sorry I know that. I don't know why I wrote y_k etc. I corrected it. Well, for y we would have 1/2 + 1/3 + 1/4 ... For x we would have 1/4+ 1/6 + 1/8...
$endgroup$
– BrainOverflow
Jan 16 at 15:14
1
1
$begingroup$
Hello and welcome to math.stackexchange. There are several issues with your question. (1) If $|a_k| ge d_k > 0$ and the series $sum_k d_k$ diverges, it does not follow that the series $sum_k a_k$ diverges. This is only true if all (or almost all) $a_k$ are positive or negative. (2) You can't write something like $y_k = sum_k a_k$, since the index $k$ is a dummy variable and the result of the series does not depend on $k$. To understand the argument in your notes/solution key, write down what it would mean for the first 4 or 6 terms.
$endgroup$
– Hans Engler
Jan 16 at 15:05
$begingroup$
Hello and welcome to math.stackexchange. There are several issues with your question. (1) If $|a_k| ge d_k > 0$ and the series $sum_k d_k$ diverges, it does not follow that the series $sum_k a_k$ diverges. This is only true if all (or almost all) $a_k$ are positive or negative. (2) You can't write something like $y_k = sum_k a_k$, since the index $k$ is a dummy variable and the result of the series does not depend on $k$. To understand the argument in your notes/solution key, write down what it would mean for the first 4 or 6 terms.
$endgroup$
– Hans Engler
Jan 16 at 15:05
$begingroup$
@HansEngler to (1) Since we only use natural Numbers I think we ignore this since it will always be positive right? (2) Yeah sorry I know that. I don't know why I wrote y_k etc. I corrected it. Well, for y we would have 1/2 + 1/3 + 1/4 ... For x we would have 1/4+ 1/6 + 1/8...
$endgroup$
– BrainOverflow
Jan 16 at 15:14
$begingroup$
@HansEngler to (1) Since we only use natural Numbers I think we ignore this since it will always be positive right? (2) Yeah sorry I know that. I don't know why I wrote y_k etc. I corrected it. Well, for y we would have 1/2 + 1/3 + 1/4 ... For x we would have 1/4+ 1/6 + 1/8...
$endgroup$
– BrainOverflow
Jan 16 at 15:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint
- $displaystylefrac{1}{2(k+1)}=frac{1}{2k+2}gefrac{1}{4k}, forall kin mathbb{N}.$
- The series $displaystylesum_{k=1}^{infty}frac{1}{4k}=frac 14 sum_{k=1}^{infty}frac{1}{k}$ is divergent.
answered Jan 16 at 15:18
mflmfl
26.9k12142
$endgroup$
$begingroup$
Thanks! I know its divergent and your solutions makes much more sense to me. But I don't get it why the university solution concludes the divergence from 1/k. Since $sum_{i=2}^infty 1/k $ > $1/2*sum_{i=2}^infty 1/k $
$endgroup$
– BrainOverflow
Jan 16 at 15:23
$begingroup$
Why do you think that? Just take $d_k=frac{1}{4k}.$
$endgroup$
– mfl
Jan 16 at 16:06
$begingroup$
I don't think that. It is the pattern solution from my university which says this. I guess there is a mistake.
$endgroup$
– BrainOverflow
Jan 16 at 16:13
add a comment |
Your Answer
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$begingroup$
Hint
- $displaystylefrac{1}{2(k+1)}=frac{1}{2k+2}gefrac{1}{4k}, forall kin mathbb{N}.$
- The series $displaystylesum_{k=1}^{infty}frac{1}{4k}=frac 14 sum_{k=1}^{infty}frac{1}{k}$ is divergent.
answered Jan 16 at 15:18
mflmfl
26.9k12142
$endgroup$
$begingroup$
Thanks! I know its divergent and your solutions makes much more sense to me. But I don't get it why the university solution concludes the divergence from 1/k. Since $sum_{i=2}^infty 1/k $ > $1/2*sum_{i=2}^infty 1/k $
$endgroup$
– BrainOverflow
Jan 16 at 15:23
$begingroup$
Why do you think that? Just take $d_k=frac{1}{4k}.$
$endgroup$
– mfl
Jan 16 at 16:06
$begingroup$
I don't think that. It is the pattern solution from my university which says this. I guess there is a mistake.
$endgroup$
– BrainOverflow
Jan 16 at 16:13
add a comment |
$begingroup$
Hint
- $displaystylefrac{1}{2(k+1)}=frac{1}{2k+2}gefrac{1}{4k}, forall kin mathbb{N}.$
- The series $displaystylesum_{k=1}^{infty}frac{1}{4k}=frac 14 sum_{k=1}^{infty}frac{1}{k}$ is divergent.
answered Jan 16 at 15:18
mflmfl
26.9k12142
$endgroup$
$begingroup$
Thanks! I know its divergent and your solutions makes much more sense to me. But I don't get it why the university solution concludes the divergence from 1/k. Since $sum_{i=2}^infty 1/k $ > $1/2*sum_{i=2}^infty 1/k $
$endgroup$
– BrainOverflow
Jan 16 at 15:23
$begingroup$
Why do you think that? Just take $d_k=frac{1}{4k}.$
$endgroup$
– mfl
Jan 16 at 16:06
$begingroup$
I don't think that. It is the pattern solution from my university which says this. I guess there is a mistake.
$endgroup$
– BrainOverflow
Jan 16 at 16:13
add a comment |
$begingroup$
Hint
- $displaystylefrac{1}{2(k+1)}=frac{1}{2k+2}gefrac{1}{4k}, forall kin mathbb{N}.$
- The series $displaystylesum_{k=1}^{infty}frac{1}{4k}=frac 14 sum_{k=1}^{infty}frac{1}{k}$ is divergent.
answered Jan 16 at 15:18
mflmfl
26.9k12142
$endgroup$
Hint
- $displaystylefrac{1}{2(k+1)}=frac{1}{2k+2}gefrac{1}{4k}, forall kin mathbb{N}.$
- The series $displaystylesum_{k=1}^{infty}frac{1}{4k}=frac 14 sum_{k=1}^{infty}frac{1}{k}$ is divergent.
answered Jan 16 at 15:18
mflmfl
26.9k12142
edited Jan 16 at 17:09
answered Jan 16 at 15:18
mflmfl
26.9k12142
answered Jan 16 at 15:18
mflmfl
26.9k12142
answered Jan 16 at 15:18
mflmfl
26.9k12142
26.9k12142
$begingroup$
Thanks! I know its divergent and your solutions makes much more sense to me. But I don't get it why the university solution concludes the divergence from 1/k. Since $sum_{i=2}^infty 1/k $ > $1/2*sum_{i=2}^infty 1/k $
$endgroup$
– BrainOverflow
Jan 16 at 15:23
$begingroup$
Why do you think that? Just take $d_k=frac{1}{4k}.$
$endgroup$
– mfl
Jan 16 at 16:06
$begingroup$
I don't think that. It is the pattern solution from my university which says this. I guess there is a mistake.
$endgroup$
– BrainOverflow
Jan 16 at 16:13
add a comment |
$begingroup$
Thanks! I know its divergent and your solutions makes much more sense to me. But I don't get it why the university solution concludes the divergence from 1/k. Since $sum_{i=2}^infty 1/k $ > $1/2*sum_{i=2}^infty 1/k $
$endgroup$
– BrainOverflow
Jan 16 at 15:23
$begingroup$
Why do you think that? Just take $d_k=frac{1}{4k}.$
$endgroup$
– mfl
Jan 16 at 16:06
$begingroup$
I don't think that. It is the pattern solution from my university which says this. I guess there is a mistake.
$endgroup$
– BrainOverflow
Jan 16 at 16:13
$begingroup$
Thanks! I know its divergent and your solutions makes much more sense to me. But I don't get it why the university solution concludes the divergence from 1/k. Since $sum_{i=2}^infty 1/k $ > $1/2*sum_{i=2}^infty 1/k $
$endgroup$
– BrainOverflow
Jan 16 at 15:23
$begingroup$
Thanks! I know its divergent and your solutions makes much more sense to me. But I don't get it why the university solution concludes the divergence from 1/k. Since $sum_{i=2}^infty 1/k $ > $1/2*sum_{i=2}^infty 1/k $
$endgroup$
– BrainOverflow
Jan 16 at 15:23
$begingroup$
Why do you think that? Just take $d_k=frac{1}{4k}.$
$endgroup$
– mfl
Jan 16 at 16:06
$begingroup$
Why do you think that? Just take $d_k=frac{1}{4k}.$
$endgroup$
– mfl
Jan 16 at 16:06
$begingroup$
I don't think that. It is the pattern solution from my university which says this. I guess there is a mistake.
$endgroup$
– BrainOverflow
Jan 16 at 16:13
$begingroup$
I don't think that. It is the pattern solution from my university which says this. I guess there is a mistake.
$endgroup$
– BrainOverflow
Jan 16 at 16:13
add a comment |
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1
$begingroup$
Hello and welcome to math.stackexchange. There are several issues with your question. (1) If $|a_k| ge d_k > 0$ and the series $sum_k d_k$ diverges, it does not follow that the series $sum_k a_k$ diverges. This is only true if all (or almost all) $a_k$ are positive or negative. (2) You can't write something like $y_k = sum_k a_k$, since the index $k$ is a dummy variable and the result of the series does not depend on $k$. To understand the argument in your notes/solution key, write down what it would mean for the first 4 or 6 terms.
$endgroup$
– Hans Engler
Jan 16 at 15:05
$begingroup$
@HansEngler to (1) Since we only use natural Numbers I think we ignore this since it will always be positive right? (2) Yeah sorry I know that. I don't know why I wrote y_k etc. I corrected it. Well, for y we would have 1/2 + 1/3 + 1/4 ... For x we would have 1/4+ 1/6 + 1/8...
$endgroup$
– BrainOverflow
Jan 16 at 15:14