Problem 1750

Let $S_n = \sum_{k=1}^{n} x_k$, $n \in \mathbb{N}^*$. Using the hypothesis, one can find a constant $c > 0$ such that $|S_n| \le c n^a$, $\forall n \in \mathbb{N}^*$. Let $n, p \in \mathbb{N}^*$; we have:
$$\begin{align*} |y_{n+p} - y_n| &= \left| \sum_{k=n+1}^{n+p} \frac{x_k}{k^b} \right| = \left| \sum_{k=n+1}^{n+p} \frac{S_k - S_{k-1}}{k^b} \right| = \\ &= \left| \frac{S_{n+p}}{(n+p+1)^b} - \frac{S_n}{(n+1)^b} + \sum_{k=n+1}^{n+p} S_k \left( \frac{1}{k^b} - \frac{1}{(k+1)^b} \right) \right| \\ &\le \frac{|S_{n+p}|}{(n+p+1)^b} + \frac{|S_n|}{(n+1)^b} + \sum_{k=n+1}^{n+p} |S_k| \left( \frac{1}{k^b} - \frac{1}{(k+1)^b} \right) \\ &\le c \left[ \frac{2}{n^{b-a}} + \sum_{k=n+1}^{n+p} k^a \left( \frac{1}{k^b} - \frac{1}{(k+1)^b} \right) \right]. \end{align*}$$
Applying the mean value theorem to the function $f(x) = x^{-\alpha}$, $x > 0$ on the interval $[i, i+1]$, with $\alpha, i > 0$, yields
$$\frac{\alpha}{(i+1)^{\alpha+1}} < \frac{1}{i^{\alpha}} - \frac{1}{(i+1)^{\alpha}} < \frac{\alpha}{i^{\alpha+1}},$$
and since $b, b-a > 0$, we obtain
$$\frac{1}{k^b} - \frac{1}{(k+1)^b} < \frac{b}{k^{b+1}} \quad \text{and} \quad \frac{b-a}{k^{b-a+1}} < \frac{1}{(k-1)^{b-a}} - \frac{1}{k^{b-a}}, \forall k \in \mathbb{N}, k \ge 2.$$
Then
$$\begin{align*} \sum_{k=n+1}^{n+p} k^a \left( \frac{1}{k^b} - \frac{1}{(k+1)^b} \right) &< \sum_{k=n+1}^{n+p} \frac{b k^a}{k^{b+1}} = \frac{b}{b-a} \sum_{k=n+1}^{n+p} \frac{b-a}{k^{b-a+1}} \\ &< \frac{b}{b-a} \sum_{k=n+1}^{n+p} \left( \frac{1}{(k-1)^{b-a}} - \frac{1}{k^{b-a}} \right) \\ &= \frac{b}{b-a} \left( \frac{1}{n^{b-a}} - \frac{1}{(n+p)^{b-a}} \right) < \frac{b}{(b-a)n^{b-a}}. \end{align*}$$
It follows that
$$|y_{n+p} - y_n| < c \left( 2 + \frac{b}{b-a} \right) \frac{1}{n^{b-a}}, \forall n, p \in \mathbb{N}^*.$$
Finally, since $\lim_{n \to \infty} n^{-(b-a)} = 0$, we obtain that $(y_n)_{n \ge 1}$ is a Cauchy sequence, hence a convergent one.