# Prove that $y_n=\frac{x_1}{1^b}+\frac{x_2}{2^b}+… +\frac{x_n}{n^b}$ is convergent

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.

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.

• 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. – Math Guy Apr 12 at 16:37
• Yes exactly. See en.wikipedia.org/wiki/Summation_by_parts . – TheSilverDoe Apr 13 at 20:15

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 $$\sum \frac 1 {n^{b+1-a}}$$.

• 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. – Kavi Rama Murthy Apr 12 at 10:10
• @user91015 My answer was quickly downvoted. That is why I made this comment. If OP wants I will surely add more details. – Kavi Rama Murthy Apr 12 at 11:42
• 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. – Math Guy Apr 12 at 16:40