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Let $a\ge 0$ and $(x_n) _{n\ge 1}$ be a sequence of real numbers. Prove that if the sequence $\left(\frac{x_1+x_2+...+x_n}{n^a} \right)_{n\ge 1}$ is bounded, then the sequence $(y_n) _{n\ge 1}$, $y_n=\frac{x_1}{1^b}+\frac{x_2}{2^b}+... +\frac{x_n}{n^b} $ is convergent $\forall b>a$.
To me, $y_n$ is reminiscent of the p-Harmonic series, but I don't know if this is actually true. Anyway, I think that we may use the Stolz-Cesaro lemma on $\frac{x_1+x_2+...+x_n}{n^a}$, but I don't know if this is of any use.

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Let's do an Abel transformation. Denoting by $S_k = x_1 + ... + x_k$ for all $k \in \mathbb{N}$, you have $$\sum_{n=1}^N \frac{x_n}{n^b} = x_1 +\sum_{n=1}^{N-1} \frac{S_{n+1}-S_n}{(n+1)^b} = x_1 +\sum_{n=1}^{N-1} \frac{S_{n+1}}{(n+1)^b} - \sum_{n=1}^{N-1} \frac{S_n}{(n+1)^b} $$

$$= x_1 +\sum_{n=2}^{N} \frac{S_n}{n^b} - \sum_{n=1}^{N-1} \frac{S_n}{(n+1)^b} =x_1 + \frac{S_N}{N^b} - \frac{x_1}{2^b} + \sum_{n=2}^{N-1} S_n \left( \frac{1}{n^b}-\frac{1}{(n+1)^b}\right) $$

Now, by hypothesis, $S_N = O(N^a)$, so because $b > a$, you have $S_N = o(N^b)$. So you have only to study the convergence of the series $$\sum_{n=2}^{N-1} S_n \left( \frac{1}{n^b}-\frac{1}{(n+1)^b}\right) $$

But one has $S_n = O(n^a)$, so $$S_n \left( \frac{1}{n^b}-\frac{1}{(n+1)^b}\right) = O \left( \frac{n^a}{n^b}-\frac{n^a}{(n+1)^b}\right)$$

Moreover, $$\frac{n^a}{n^b}-\frac{n^a}{(n+1)^b} = \frac{n^a(n+1)^b - n^an^b}{n^b(n+1)^b} \sim \frac{n^a}{(n+1)^b} \left( \left(1+ \frac{1}{n} \right)^b-1\right)$$

$$ \sim \frac{n^a}{(n+1)^b} \left( \frac{b}{n} \right) \sim \frac{b}{n^{b-a+1}}$$

This last series converges, so you deduce that the series of general term $$ S_n \left( \frac{1}{n^b}-\frac{1}{(n+1)^b}\right)$$

also converges, so you deduce that your original series converges.

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  • $\begingroup$ The Abel transformation is basically writing $x_n=S_{n+1}-S_n$ and separating the sums? I am asking because I have never seen this technique before. $\endgroup$ – Math Guy Apr 12 at 16:37
  • $\begingroup$ Yes exactly. See en.wikipedia.org/wiki/Summation_by_parts . $\endgroup$ – TheSilverDoe Apr 13 at 20:15
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Hints: let $s_n=a_1+a_2+...+a_n$. Write $\sum \frac {x_n} {n^{b}}$ as $\sum \frac {s_n-s_{n-1}} {n^{b}}$ and re-write this in terms of $\sum s_n \big({\frac 1 {n^{b}} -\frac 1 {(n+1)^{b}}\big)}$. Using the hypothesis we see that it remains only to show the convergence of $\sum \big({\frac {n^{a}} {n^{b}} -\frac {n^{a}} {(n+1)^{b}}\big)}$. Compare this with the series $\sum \frac 1 {n^{b+1-a}}$.

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  • $\begingroup$ This is the standard technique of 'summation by parts' and the only natural way to answer this question. If more details are need I can include them. $\endgroup$ – Kavi Rama Murthy Apr 12 at 10:10
  • $\begingroup$ @user91015 My answer was quickly downvoted. That is why I made this comment. If OP wants I will surely add more details. $\endgroup$ – Kavi Rama Murthy Apr 12 at 11:42
  • $\begingroup$ It wasn't downvoted by me, some other user downvoted it, I actually upvoted your work. In the mantime, another user posted a full solution using your hint. $\endgroup$ – Math Guy Apr 12 at 16:40

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