Assume that we have a positive sequence $\{a_n\}$ with its Cesàro mean converges: $$\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^{N}a_n<\infty\,.$$ I am wondering if the following summation converges: $$\lim_{N\to\infty}\sum_{n=1}^{N}\frac{a_n}{n^2}$$

Any proof is appreciated!

  • $\begingroup$ Squeeze theorem? Would this work? $\endgroup$ – Jihoon Kang Sep 26 '17 at 15:43
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    $\begingroup$ Do you know summation by parts? $\endgroup$ – Daniel Fischer Sep 26 '17 at 15:52
  • $\begingroup$ @JihoonKang Sorry but how? $\endgroup$ – Did Sep 26 '17 at 17:03
  • $\begingroup$ @Did I was (probably) wrong - I thought it might be possible without thinking through the details. $\endgroup$ – Jihoon Kang Sep 27 '17 at 1:11
  • $\begingroup$ @JihoonKang May I suggest to avoid at all cost such random comments? They are a (minor) plague of the site. $\endgroup$ – Did Sep 27 '17 at 8:10

Define $s_0=0, s_n = a_1 + \cdots + a_n, n>0.$ Summing by parts (note Daniel Fischer mentioned this in a comment) gives

$$ \sum_{k=1}^{n}\frac{a_k}{k^2}= \sum_{k=1}^{n}\frac{s_k-s_{k-1}}{k^2}= \frac{s_n}{n^2} + \sum_{k=1}^{n-1}s_k\cdot \left ( \frac{1}{k^2}-\frac{1}{(k+1)^2}\right)$$

The first term on the right $\to 0,$ so we can ignore it. The remaining sum equals

$$\sum_{k=1}^{n-1}s_k\cdot \frac{2k+1}{k^2(k+1)^2}= \sum_{k=1}^{n-1}\frac{s_k}{k}\cdot \frac{2k+1}{k(k+1)^2}.$$

The sequence $\dfrac{s_k}{k}$ is bounded. Since $\sum_{k=1}^{\infty}\dfrac{2k+1}{k(k+1)^2} < \infty$ we have

$$\sum_{k=1}^{\infty}\frac{s_k}{k}\cdot \frac{2k+1}{k(k+1)^2}<\infty.$$ This implies $\sum_{k=1}^{\infty}\dfrac{a_k}{k^2}<\infty.$

  • $\begingroup$ Clear and concise. (+1) $\endgroup$ – Mark Viola Sep 26 '17 at 17:57
  • $\begingroup$ @MarkViola Thank you sir. $\endgroup$ – zhw. Sep 26 '17 at 18:24

Given any real sequence $(a_n)_{n\ge 1}$, construct an auxillary sequence $(b_n)_{n\ge 0}$ by

$$b_n \stackrel{def}{=}\begin{cases}\frac1n\sum\limits_{k=1}^n a_k, & n > 0\\0, & n = 0\end{cases}$$ Whenever $b_n$ is bounded, i.e there is a $M > 0$ such that $|b_n| < M$ for all $n$, we can decompose the sum at hand into two pieces:

$$\sum_{n=1}^\infty \frac{a_n}{n^2} = \sum_{n=1}^\infty \frac{nb_n - (n-1)b_{n-1}}{n^2} = \sum_{n=1}^\infty \left(\frac{b_n}{n^2} + \frac{n-1}{n^2}(b_n - b_{n-1})\right) $$ The first piece $\displaystyle\;\sum_{n=1}^\infty \frac{b_n}{n^2}\;$ will be absolutely convergent by comparision test against $\displaystyle\;\sum_{n=1}^\infty\frac{M}{n^2}$.

The second piece $\displaystyle\;\sum_{n=1}^\infty \frac{(n-1)}{n^2} (b_n - b_{n-1})$ will be conditionally convergent by Dirichlet's test because $\displaystyle\;\frac{n-1}{n^2}\;$ is monontonic decreasing to zero for $n > 1$ and $\displaystyle\;\left|\sum_{n=1}^N (b_n - b_{n-1})\right| = |b_N| < M\;$ is bounded.

We don't really need $(a_n)$ be positive nor $(b_n)$ converges. As long as $b_n$ is bounded, the sum at hand converges.

  • $\begingroup$ Thank you so much for the proof. One question: it seems that when $n=1$, $\frac{n-1}{n^2}=0$ and so $\frac{n-1}{n^2}$ is not monotonic decreasing on $[1,\,\infty)$. Will this invalidate the proof? Thx $\endgroup$ – user99015 Sep 26 '17 at 16:57
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    $\begingroup$ @user99015 it won't invalidate the proof, the test remains to work if finitely many terms are missing the conditions. $\endgroup$ – achille hui Sep 26 '17 at 17:01
  • $\begingroup$ Nicely done! (+1) $\endgroup$ – Mark Viola Sep 26 '17 at 17:57

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