Finding the convergence of a sequence $\sum_{k=1}^n\frac{x_k}{k^b}$

Question

I found a problem that had was giving a sequence $x_n$ of real numbers and the following sequence $$\frac{\sum_{k=1}^n{x_k}}{n^a}$$ is bounded for any $a > 0$. Given $b > a$ the problem asked to prove that the sequence $$\sum_{k=1}^n\frac{x_k}{k^b}$$ is convergent. I will present the way I approached this problem. Firstly, let's denote $s_n = \sum_{k=1}^nx_k$ the sequence of partial sums. We know that $\frac{s_n}{n^a}$ is bounded so there exists $M>0$ such that $|s_n|<M|n^a|$. Because $n$ is a natural number we get that $|s_n|<Mn^a$. To prove that $y_n$ converges I thought to try and prove that $y_n$ is a Cauchy sequence($\mathbb{R}$ is a complete metric space). So $$|y_{n+p}-y_n|=\Big|\sum_{k=n+1}^{n+p}\frac{s_k-s_{k-1}}{k^b}\Big|=\Big|\sum_{k=n+1}^{n+p}\frac{s_k}{k^b}-\sum_{k=n+1}^{n+p}\frac{s_{k-1}}{k^b}\Big|$$ Using the triangle inequality we can see that $$|y_{n+p}-y_n|\le\Big|\sum_{k=n+1}^{n+p}\frac{s_k}{k^b}\Big|+\Big|\sum_{k=n+1}^{n+p}\frac{s_{k-1}}{k^b}\Big|\le\Big|\sum_{k=n+1}^{n+p}\frac{Mn^a}{k^b}\Big|+\Big|\sum_{k=n+1}^{n+p}\frac{Mn^a}{k^b}\Big|=2Mn^a\Big|\sum_{k=n+1}^{n+p}\frac{1}{k^b}\Big|$$ The question I have is if it is rigurous to say that if $$n^a\sum_{k=n+1}^{n+p}\frac{1}{k^b}$$ converges to $0$ when $n -> \infty$ then $y_n$ is a Cauchy sequence. Isn't $a_n$ a Cauchy sequence if $|a_n-a_m|<\epsilon$ as $m$ and $n$ independently approach infinity? I also tried to find the limit of the last thing sum and i got $$n^a\sum_{k=n+1}^{n+p}\frac{1}{k^b}=\frac{\sum_{k=n+1}^{n+p}\frac{1}{k^b}}{n^{-a}}$$ Since $n^{-a}$ approaches $0$ as $n$ approaches infinity and it is strictly decreasing we can use the Stolz-Cesaro Theorem. $$\frac{\frac{1}{(n+p+1)^b}-\frac{1}{(n+1)^b}}{\frac{1}{(n+1)^a}-\frac{1}{n^a}}$$ If the last limit exists then our original limit has the same limit as this sequence. I put this in wolfram alpha with values of $a=2$ and $b=3$ and $a=2.99, b=3$ and the limit was $0$. I tried to compute the limit by hand, I even tried to consider $n$ a real number and use l'Hopital's Rule but I didn't get that far. Any ideas on how I should continue?

Answer 1

Your using the Abel transformation (i.e. integration by parts) is the right idea. $$\sum_{k=1}^n\frac{x_k}{k^b}=\sum_{k=1}^n\frac{s_k-s_{k-1}}{k^b}=\frac{s_n}{n^b}+\sum_{k=1}^{n-1}s_k\left( \frac 1 {k^b}-\frac 1 {(k+1)^b}\right)\tag{1}$$ Now, by definition, there exists $M>0$ such that for all $n$, we have $\left|\frac {s_n}{n^a}\right|\leq M$. Thus: $$\left|\frac{s_n}{n^b}\right|\leq \frac{M}{n^{b-a}}$$ Since $b>a$, this implies that the first term in $(1)$, i.e. $\frac{s_n}{n^b}$, converges to $0$.

Now for the second term, the sum, notice that $$\left| \frac 1 {k^b}-\frac 1 {(k+1)^b}\right|=\frac{(k+1)^b-k^b}{k^b(k+1)^b}\leq \frac{b(k+1)^{b-1}}{k^b (k+1)^b}=\frac{b}{k^b(k+1)}$$ Thus $$\sum_{k=1}^{n-1}\left|s_k\left( \frac 1 {k^b}-\frac 1 {(k+1)^b}\right)\right| \leq b\sum_{k=1}^{n-1}\frac{|s_k|}{k^a}\frac{1}{k^{b-a}(k+1)}\leq bM\sum_{k=1}^{n-1}\frac{1}{k^{b-a}(k+1)}$$ and since $b>a$, the sum on the right-hand side converges (each term behaves like $\frac 1 {k^{1+b-a}}$), which implies the left-hand side converges.

Going back to $(1)$, the first term converges to $0$, and the second term converges (in fact, absolutely), so the whole thing converges.