Question 25
Let $a \ge 0$ and let $(x_n)_{n \ge 1}$ be a sequence of real numbers. Given that $\left(\frac{x_1+\dots+x_n}{n^a}\right)_{n \ge 1}$ is a bounded sequence, prove that the sequence $(y_n)_{n \ge 1}$, defined by
$$
y_n = \frac{x_1}{1^b} + \frac{x_2}{2^b} + \dots + \frac{x_n}{n^b},
$$
is a convergent sequence for all $b \gt a$.
$$
y_n = \frac{x_1}{1^b} + \frac{x_2}{2^b} + \dots + \frac{x_n}{n^b},
$$
is a convergent sequence for all $b \gt a$.