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If $sum_n=1^infty a_n$ converges, prove that $lim_nto infty (1/n) sum_k=1^n ka_k = 0$.

Show that if $sum limits_n=1^infty a_n$ converges then $limlimits_ntoinftyfrac1nsum _k=1^n ka_k=0$If the series $sum_k=1^infty a_k$ converges, then we have $dfracsum_k=1^n k a_kn$ goes to $0$ as $n$ goes to $infty$Testing if a geometric series converges by taking limit to infinityFinding Limit of Sequence (x^x)/ (x!)Examine the convergence of $sum_n=1^infty frac1n(log_2n)$Convergence of $sum_n = 1^infty 1/n^2$.If the positive series $sum a_n$ converges does $sum a_n log(a_n)$ converge?Prove that $sum_limitsk=0^inftyfraccos^2kk+1$ divergesHow can we conclude that this series converges faster?If the series $a_n$ converges to $L in (-1,1)$ then the series $a_n^n$ converges to 0Determine whether $sum_n=1^infty frac sin(n)n$ convergesKnowing that $lim_nto infty a_n = 0$ and that $b_n$ is bounded, find $lim_nto infty (a_n b_n)$

1

1

$begingroup$

the series $a_n$ is any arbitrary converging series.

My thought process was that the $1/n$ will definitely go to zero as n approaches infinity; however, the series $k*a_k$ seems to approach infinity at the same rate. This confuses me because if both the numerator and denominator approach infinity, I think that the overall equation will equal 1, not zero.

I also considered that the sum is geometric, but I do not believe that is the case, because both $a_k$ and $k$ are variables that are not constant. I was thinking that I could argue if the series $a_k$ converges to a point, eventually the terms of the sequence will be so close to each other that it resembles a constant multiplier, but that argument is not very strong.

Any tips, hints, or leads at the solution are appreciated!

sequences-and-series convergence

share|cite|improve this question

edited Oct 11 '15 at 18:22

zhw.

74.8k43275

asked Oct 11 '15 at 18:11

Kevin McDonoughKevin McDonough

31817

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is $sum_n=1^inftya_n$?
  $endgroup$
  – Oria Gruber
  Oct 11 '15 at 18:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  it is an arbitrary sequence that converges
  $endgroup$
  – Kevin McDonough
  Oct 11 '15 at 18:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And where does $x$ come in? You meant $n$ I suppose.
  $endgroup$
  – Oria Gruber
  Oct 11 '15 at 18:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  you're right, I'm new to MathJax
  $endgroup$
  – Kevin McDonough
  Oct 11 '15 at 18:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And you also meant $k*a_k$? Because $lim_n to 0 a_n$ doesnt make much sense. I'm not trying to be rude, I'm just trying to understand the question so I could maybe help.
  $endgroup$
  – Oria Gruber
  Oct 11 '15 at 18:15
  
  

  
  
  
  

 | 
show 3 more comments

1

1

$begingroup$

the series $a_n$ is any arbitrary converging series.

My thought process was that the $1/n$ will definitely go to zero as n approaches infinity; however, the series $k*a_k$ seems to approach infinity at the same rate. This confuses me because if both the numerator and denominator approach infinity, I think that the overall equation will equal 1, not zero.

I also considered that the sum is geometric, but I do not believe that is the case, because both $a_k$ and $k$ are variables that are not constant. I was thinking that I could argue if the series $a_k$ converges to a point, eventually the terms of the sequence will be so close to each other that it resembles a constant multiplier, but that argument is not very strong.

Any tips, hints, or leads at the solution are appreciated!

sequences-and-series convergence

share|cite|improve this question

edited Oct 11 '15 at 18:22

zhw.

74.8k43275

asked Oct 11 '15 at 18:11

Kevin McDonoughKevin McDonough

31817

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is $sum_n=1^inftya_n$?
  $endgroup$
  – Oria Gruber
  Oct 11 '15 at 18:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  it is an arbitrary sequence that converges
  $endgroup$
  – Kevin McDonough
  Oct 11 '15 at 18:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And where does $x$ come in? You meant $n$ I suppose.
  $endgroup$
  – Oria Gruber
  Oct 11 '15 at 18:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  you're right, I'm new to MathJax
  $endgroup$
  – Kevin McDonough
  Oct 11 '15 at 18:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And you also meant $k*a_k$? Because $lim_n to 0 a_n$ doesnt make much sense. I'm not trying to be rude, I'm just trying to understand the question so I could maybe help.
  $endgroup$
  – Oria Gruber
  Oct 11 '15 at 18:15
  
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

the series $a_n$ is any arbitrary converging series.

My thought process was that the $1/n$ will definitely go to zero as n approaches infinity; however, the series $k*a_k$ seems to approach infinity at the same rate. This confuses me because if both the numerator and denominator approach infinity, I think that the overall equation will equal 1, not zero.

I also considered that the sum is geometric, but I do not believe that is the case, because both $a_k$ and $k$ are variables that are not constant. I was thinking that I could argue if the series $a_k$ converges to a point, eventually the terms of the sequence will be so close to each other that it resembles a constant multiplier, but that argument is not very strong.

Any tips, hints, or leads at the solution are appreciated!

sequences-and-series convergence

share|cite|improve this question

edited Oct 11 '15 at 18:22

zhw.

74.8k43275

asked Oct 11 '15 at 18:11

Kevin McDonoughKevin McDonough

31817

$endgroup$

share|cite|improve this question

edited Oct 11 '15 at 18:22

zhw.

74.8k43275

asked Oct 11 '15 at 18:11

Kevin McDonoughKevin McDonough

31817

share|cite|improve this question

edited Oct 11 '15 at 18:22

zhw.

74.8k43275

asked Oct 11 '15 at 18:11

Kevin McDonoughKevin McDonough

31817

edited Oct 11 '15 at 18:22

zhw.

74.8k43275

edited Oct 11 '15 at 18:22

zhw.

74.8k43275

asked Oct 11 '15 at 18:11

Kevin McDonoughKevin McDonough

31817

asked Oct 11 '15 at 18:11

Kevin McDonoughKevin McDonough

31817

3 Answers
3

active

oldest

votes

3

$begingroup$

Let $x_n=sum_k=1^n a_k$. The hypothesis is that $x_n$ converges to some $x$. Note that $sum_i=1^n x_i = sum_k=1^n (n-k+1)a_k$, and recall (or if necessary, prove) that if $y_nto y$ then $$hat y_n=frac1nsum_i=1^n y_ito y$$

share|cite|improve this answer

edited Oct 11 '15 at 18:44

answered Oct 11 '15 at 18:19

Pedro Tamaroff♦Pedro Tamaroff

97.5k10153299

$endgroup$

add a comment | 

1

$begingroup$

Let $S_k = sum_j=1^ka_j.$ We are given $S_k to$ some $L.$ Summing by parts, we get

$$sum_k=1^nka_k = nS_n - sum_k=1^n-1S_k.$$
Dividing by $n$ gives $S_n$ minus $(n-1)/n$ times the $(n-1)$st Cesaro mean of the $S_k,$ which also converges to $L.$ Our limit is therefore $L-L=0.$

share|cite|improve this answer

answered Oct 11 '15 at 18:44

zhw.zhw.

74.8k43275

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You had an incorrect solution, which you deleted. I didn't see your new solution, which was posted 27 seconds before mine. I think I'll leave mine up.
  $endgroup$
  – zhw.
  Oct 11 '15 at 18:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hehe, no problem.
  $endgroup$
  – Pedro Tamaroff♦
  Oct 11 '15 at 18:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (The solution had a typo, to be fair =) )
  $endgroup$
  – Pedro Tamaroff♦
  Oct 11 '15 at 19:00
  

  
  
  
  

add a comment | 

1

$begingroup$

Hint: from the convergence of $sum_ngeq 1a_n$, we know that the sequence $a_n_ngeq 1$ has bounded partial sums. We have:

$$ frac1nsum_k=1^n k,a_k = sum_k=1^nfrackn,a_k = sum_k=1^n a_k - sum_k=1^nleft(1-fracknright)a_k. $$
Now apply Dirichlet's criterion to the last sum regarded as a series.

Once you know that $frac1nsum_k=1^n k,a_k$ is convergent, convergence to zero follows from the dominated convergence theorem.

share|cite|improve this answer

edited Oct 11 '15 at 19:34

answered Oct 11 '15 at 18:18

Jack D'AurizioJack D'Aurizio

292k33284672

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not convinced Dirichlet is applicable here, since you do not know that $a_n geq a_n+1$.
  $endgroup$
  – Oria Gruber
  Oct 11 '15 at 18:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @OriaGruber: the other way. $a_n_ngeq 1$ has bounded partial sums and $left(1-k/nright)$ is decreasing to zero.
  $endgroup$
  – Jack D'Aurizio
  Oct 11 '15 at 18:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That only shows convergence - not convergence to 0.
  $endgroup$
  – A.S.
  Oct 11 '15 at 18:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @A.S.: once you know that convergence holds, the other part is easy.
  $endgroup$
  – Jack D'Aurizio
  Oct 11 '15 at 19:10
  

  
  
  
  

add a comment | 

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3

$begingroup$

Let $x_n=sum_k=1^n a_k$. The hypothesis is that $x_n$ converges to some $x$. Note that $sum_i=1^n x_i = sum_k=1^n (n-k+1)a_k$, and recall (or if necessary, prove) that if $y_nto y$ then $$hat y_n=frac1nsum_i=1^n y_ito y$$

share|cite|improve this answer

edited Oct 11 '15 at 18:44

answered Oct 11 '15 at 18:19

Pedro Tamaroff♦Pedro Tamaroff

97.5k10153299

$endgroup$

add a comment | 

3

$begingroup$

Let $x_n=sum_k=1^n a_k$. The hypothesis is that $x_n$ converges to some $x$. Note that $sum_i=1^n x_i = sum_k=1^n (n-k+1)a_k$, and recall (or if necessary, prove) that if $y_nto y$ then $$hat y_n=frac1nsum_i=1^n y_ito y$$

share|cite|improve this answer

edited Oct 11 '15 at 18:44

answered Oct 11 '15 at 18:19

Pedro Tamaroff♦Pedro Tamaroff

97.5k10153299

$endgroup$

add a comment | 

$begingroup$

Let $x_n=sum_k=1^n a_k$. The hypothesis is that $x_n$ converges to some $x$. Note that $sum_i=1^n x_i = sum_k=1^n (n-k+1)a_k$, and recall (or if necessary, prove) that if $y_nto y$ then $$hat y_n=frac1nsum_i=1^n y_ito y$$

share|cite|improve this answer

edited Oct 11 '15 at 18:44

answered Oct 11 '15 at 18:19

Pedro Tamaroff♦Pedro Tamaroff

97.5k10153299

$endgroup$

share|cite|improve this answer

edited Oct 11 '15 at 18:44

answered Oct 11 '15 at 18:19

Pedro Tamaroff♦Pedro Tamaroff

97.5k10153299

answered Oct 11 '15 at 18:19

Pedro Tamaroff♦Pedro Tamaroff

97.5k10153299

answered Oct 11 '15 at 18:19

Pedro Tamaroff♦Pedro Tamaroff

97.5k10153299

1

$begingroup$

Let $S_k = sum_j=1^ka_j.$ We are given $S_k to$ some $L.$ Summing by parts, we get

$$sum_k=1^nka_k = nS_n - sum_k=1^n-1S_k.$$
Dividing by $n$ gives $S_n$ minus $(n-1)/n$ times the $(n-1)$st Cesaro mean of the $S_k,$ which also converges to $L.$ Our limit is therefore $L-L=0.$

share|cite|improve this answer

answered Oct 11 '15 at 18:44

zhw.zhw.

74.8k43275

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You had an incorrect solution, which you deleted. I didn't see your new solution, which was posted 27 seconds before mine. I think I'll leave mine up.
  $endgroup$
  – zhw.
  Oct 11 '15 at 18:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hehe, no problem.
  $endgroup$
  – Pedro Tamaroff♦
  Oct 11 '15 at 18:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (The solution had a typo, to be fair =) )
  $endgroup$
  – Pedro Tamaroff♦
  Oct 11 '15 at 19:00
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $S_k = sum_j=1^ka_j.$ We are given $S_k to$ some $L.$ Summing by parts, we get

$$sum_k=1^nka_k = nS_n - sum_k=1^n-1S_k.$$
Dividing by $n$ gives $S_n$ minus $(n-1)/n$ times the $(n-1)$st Cesaro mean of the $S_k,$ which also converges to $L.$ Our limit is therefore $L-L=0.$

share|cite|improve this answer

answered Oct 11 '15 at 18:44

zhw.zhw.

74.8k43275

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You had an incorrect solution, which you deleted. I didn't see your new solution, which was posted 27 seconds before mine. I think I'll leave mine up.
  $endgroup$
  – zhw.
  Oct 11 '15 at 18:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hehe, no problem.
  $endgroup$
  – Pedro Tamaroff♦
  Oct 11 '15 at 18:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (The solution had a typo, to be fair =) )
  $endgroup$
  – Pedro Tamaroff♦
  Oct 11 '15 at 19:00
  

  
  
  
  

add a comment | 

$begingroup$

Let $S_k = sum_j=1^ka_j.$ We are given $S_k to$ some $L.$ Summing by parts, we get

$$sum_k=1^nka_k = nS_n - sum_k=1^n-1S_k.$$
Dividing by $n$ gives $S_n$ minus $(n-1)/n$ times the $(n-1)$st Cesaro mean of the $S_k,$ which also converges to $L.$ Our limit is therefore $L-L=0.$

share|cite|improve this answer

answered Oct 11 '15 at 18:44

zhw.zhw.

74.8k43275

$endgroup$

share|cite|improve this answer

answered Oct 11 '15 at 18:44

zhw.zhw.

74.8k43275

answered Oct 11 '15 at 18:44

zhw.zhw.

74.8k43275

answered Oct 11 '15 at 18:44

zhw.zhw.

74.8k43275

1

$begingroup$

Hint: from the convergence of $sum_ngeq 1a_n$, we know that the sequence $a_n_ngeq 1$ has bounded partial sums. We have:

$$ frac1nsum_k=1^n k,a_k = sum_k=1^nfrackn,a_k = sum_k=1^n a_k - sum_k=1^nleft(1-fracknright)a_k. $$
Now apply Dirichlet's criterion to the last sum regarded as a series.

Once you know that $frac1nsum_k=1^n k,a_k$ is convergent, convergence to zero follows from the dominated convergence theorem.

share|cite|improve this answer

edited Oct 11 '15 at 19:34

answered Oct 11 '15 at 18:18

Jack D'AurizioJack D'Aurizio

292k33284672

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not convinced Dirichlet is applicable here, since you do not know that $a_n geq a_n+1$.
  $endgroup$
  – Oria Gruber
  Oct 11 '15 at 18:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @OriaGruber: the other way. $a_n_ngeq 1$ has bounded partial sums and $left(1-k/nright)$ is decreasing to zero.
  $endgroup$
  – Jack D'Aurizio
  Oct 11 '15 at 18:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That only shows convergence - not convergence to 0.
  $endgroup$
  – A.S.
  Oct 11 '15 at 18:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @A.S.: once you know that convergence holds, the other part is easy.
  $endgroup$
  – Jack D'Aurizio
  Oct 11 '15 at 19:10
  

  
  
  
  

add a comment | 

1

$begingroup$

Hint: from the convergence of $sum_ngeq 1a_n$, we know that the sequence $a_n_ngeq 1$ has bounded partial sums. We have:

$$ frac1nsum_k=1^n k,a_k = sum_k=1^nfrackn,a_k = sum_k=1^n a_k - sum_k=1^nleft(1-fracknright)a_k. $$
Now apply Dirichlet's criterion to the last sum regarded as a series.

Once you know that $frac1nsum_k=1^n k,a_k$ is convergent, convergence to zero follows from the dominated convergence theorem.

share|cite|improve this answer

edited Oct 11 '15 at 19:34

answered Oct 11 '15 at 18:18

Jack D'AurizioJack D'Aurizio

292k33284672

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not convinced Dirichlet is applicable here, since you do not know that $a_n geq a_n+1$.
  $endgroup$
  – Oria Gruber
  Oct 11 '15 at 18:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @OriaGruber: the other way. $a_n_ngeq 1$ has bounded partial sums and $left(1-k/nright)$ is decreasing to zero.
  $endgroup$
  – Jack D'Aurizio
  Oct 11 '15 at 18:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That only shows convergence - not convergence to 0.
  $endgroup$
  – A.S.
  Oct 11 '15 at 18:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @A.S.: once you know that convergence holds, the other part is easy.
  $endgroup$
  – Jack D'Aurizio
  Oct 11 '15 at 19:10
  

  
  
  
  

add a comment | 

$begingroup$

Hint: from the convergence of $sum_ngeq 1a_n$, we know that the sequence $a_n_ngeq 1$ has bounded partial sums. We have:

$$ frac1nsum_k=1^n k,a_k = sum_k=1^nfrackn,a_k = sum_k=1^n a_k - sum_k=1^nleft(1-fracknright)a_k. $$
Now apply Dirichlet's criterion to the last sum regarded as a series.

Once you know that $frac1nsum_k=1^n k,a_k$ is convergent, convergence to zero follows from the dominated convergence theorem.

share|cite|improve this answer

edited Oct 11 '15 at 19:34

answered Oct 11 '15 at 18:18

Jack D'AurizioJack D'Aurizio

292k33284672

$endgroup$

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edited Oct 11 '15 at 19:34

answered Oct 11 '15 at 18:18

Jack D'AurizioJack D'Aurizio

292k33284672

answered Oct 11 '15 at 18:18

Jack D'AurizioJack D'Aurizio

292k33284672

answered Oct 11 '15 at 18:18

Jack D'AurizioJack D'Aurizio

292k33284672