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Let $a_k(n)$ be such that $\sum_k a_k(n)$ converges for all $n$ and $a_k(n)\to0$ as $n\to\infty$ for all $k$. Then I think it holds that $\sum_k a_k(n)\to0$ as $n\to\infty$.

For example, if $\displaystyle\sum_{k=1}^\infty a_k$ is any convergent series then $\displaystyle\sum_{k=1}^\infty a_k(n)=\sum_{k=1}^\infty \frac{a_k}{n}=\frac1n\sum_{k=1}^\infty a_k\to0.$

I guess one should write $a_k(n)=\delta(n)\varepsilon(k)$ where $\delta(n)$ and $\varepsilon(k)$ respectively don't depend on $k$ and $n$... then one gets $\sum_k a_k(n)=\delta(n)\sum_k\varepsilon(k)\to0$ as $n\to\infty$, but how can we be sure that such $\delta$ and $\varepsilon$ exist?

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  • $\begingroup$ I don't think such $\delta,\epsilon$ can exists. Think of powers and sums $k,n$ mixed under a root sign wrapped by a sin/cos function. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 15 '17 at 21:10
  • $\begingroup$ @GNUSupporter In general they don't have to be elementary functions, but they may exist nonetheless $\endgroup$ – Richard Dec 15 '17 at 21:14
  • $\begingroup$ My point is, they may be mixed up inside the argument of some (non)elementary functions so that we can't (or we don't know how to) separate them. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 15 '17 at 21:17
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Consider $a_k(n) = \begin{cases}2^{n-k} & \text{if} \ k \ge n \\ 0 & \text{if} \ k < n\end{cases}$.

Then, for any $k$, we have $a_k(n) = 0$ for all $n > k$, and thus, $\displaystyle\lim_{n \to \infty}a_k(n) = 0$.

Also, for any $n$, we have $\displaystyle\sum_{k = 0}^{\infty}a_k(n) = \sum_{k = 0}^{n-1}0 + \sum_{k = n}^{\infty}2^{n-k} = 0 + 2 = 2$.

So, $\displaystyle\sum_{k = 0}^{\infty}a_k(n)$ converges for all $k$, but $\displaystyle\lim_{n \to \infty}\sum_{k = 0}^{\infty}a_k(n) = 2 \neq 0$.

To make the conclusion hold, you will probably need some sort of monotonicity or uniformity condition.

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  • $\begingroup$ Thanks to you and all the rest! May I ask you: what if we require $\lim\limits_{n\to\infty}\lvert a_j(n) -a_k(n)\rvert=0$ for all $j,k$? $\endgroup$ – Richard Dec 15 '17 at 22:11
  • $\begingroup$ @Richard that's already the case for this example; note that for any given $j,k$ then for $n\gt j,k$ the quantity in the absolute-value brackets is identically zero and so certainly its limit as $n\to\infty$ is zero. $\endgroup$ – Steven Stadnicki Dec 15 '17 at 22:15
  • $\begingroup$ @StevenStadnicki Darn it, I had confused $k$ and $n$ :) thanks. $\endgroup$ – Richard Dec 15 '17 at 22:18
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    $\begingroup$ You have to admit that @zhw's solution beats both of ours for simplicity. :) $\endgroup$ – John Hughes Dec 15 '17 at 22:38
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    $\begingroup$ @Richard counterexample with uniform convergence to $0$ $$\begin{matrix}1&0&0&0&0&0&\dots\\ 0&1/2&1/2&0&0&0&\dots&\\0&0&0&1/3&1/3&1/3&\dots &\\.&.&.&.&.&.& \dots \end{matrix} $$ $\endgroup$ – zhw. Dec 16 '17 at 19:15
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$1, 0,0,0,\dots$

$0,1, 0,0,0,\dots $

$0,0,1, 0,0,0,\dots $

$0,0,0,1, 0,0,0,\dots $

$\dots$

$\dots$

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No. If I follow your question correctly, that is definitely not a valid conclusion. Here's an example:

Define $a_k(n) = 0$ if $n \le k$, but $a_k(n) = \frac{1}{2^{n-k}}$ otherwise.

Then clearly $a_k(n) \to 0$ for all $k$ as $n \to \infty$.

And the sum $\sum_k a_k(n)$ is just 1.

Hence $\lim_{n \to \infty} \sum_k a_k(n)$ is $1$, not zero.

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  • $\begingroup$ The question is asking if $\sum_{k}a_k(n) \to 0$ as $n \to \infty$ not $k$. $\endgroup$ – JimmyK4542 Dec 15 '17 at 21:18
  • $\begingroup$ Yeah...I just realized that and am editing as we speak. :) $\endgroup$ – John Hughes Dec 15 '17 at 21:19
  • $\begingroup$ Fixed (apparently concurrently with you:) ). $\endgroup$ – John Hughes Dec 15 '17 at 21:29
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If you consider $\zeta(n)=\sum_{k=1}^{\infty}\frac{1}{k^n}$, as $n$ gets large, the sequence $\langle\frac{1}{k^n}\rangle_{_n}\rightarrow0$ for all $k$ but the partial sums tend to $1$. While it may hold for certain sums it surely isn't applicable for all.

addendum: here I am taking $a_k(n)=\frac{1}{k^n}$

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  • $\begingroup$ Your formula for $\zeta(n)$ is wrong (you don't want the $k=0$ term), and the fact that the partial sums tend to 1 is because the $k=1$ term is uniformly 1 for all $n$, so the hypothesis in the OP (that $a_k(n)\to0$ for all $k$) isn't met. $\endgroup$ – Steven Stadnicki Dec 16 '17 at 2:00
  • $\begingroup$ My bad. Correcting it right now $\endgroup$ – Jepsilon Dec 16 '17 at 10:25

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