I am working on the following question:
If $a_{k}\geq0$ is a bounded sequence, prove that
$\sum_{k=0}^{\infty} \frac {a_{k}} {(k+1)^{p}}$
converges for all $p>1.$
I will begin with what I know and then show what steps I have taken towards a proof. I recognized that
$\sum_{k=0}^{\infty}\frac {1} {(k+1)^{p}} \leq \sum_{k=1}^{\infty}\frac {1} {k^{p}}$
and that both sides of the inequality are convergent. The right side is a p-series with $p>1$ and as such is convergent. By the comparison test the left side of the inequality is also convergent. I also know that $a_{k}$ is bounded so there is a number $m \geq 0$ and a number $M<\infty$ such that
$m\leq a_{k} \leq M.$
From here I made the following steps
$m\cdot \frac {1} {k^{p}} \leq \frac {a_{k}} {k^{p}} \leq M\cdot \frac {1} {k^{p}}$
$m\cdot \sum_{k=1}^{\infty} \frac {1} {k^{p}} \leq \sum_{k=1}^{\infty} \frac {a_{k}} {k^{p}} \leq M\cdot \sum_{k=1}^{\infty} \frac {1} {k^{p}}$
Which I beleive shows that $\sum_{k=1}^{\infty} \frac {a_{k}} {k^{p}}$ is convergent. This results in
$\sum_{k=0}^{\infty} \frac {a_{k}} {(k+1)^{p}}\leq \sum_{k=1}^{\infty} \frac {a_{k}} {k^{p}},$
showing that the sequence is convergent for all $p>1$. I was wondering if my reasoning is sound, and if there is something I should pay particular attention to when writing the proof.