# Bounded Sequence Proof

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.

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Comparison test, if $|a_n|<M$, then the absolute value of the $n$-th term of our sequence is $<\frac{M}{n^p}$. –  André Nicolas Mar 15 '12 at 1:58

What you’ve done is basically fine, though you’ve over-complicated it a bit. In particular, there’s no need to bother with the lower bound $m$, since you know that $0$ is a lower bound and don’t care about more than that. All you really need is
$$\sum_{k\ge 1}\frac{a_k}{(k+1)^p}\le M\sum_{k\ge 1}\frac1{k^p}\;,$$
justified on the grounds that for every term you have $$0\le\frac{a_k}{(k+1)^p}\le\frac{M}{(k+1)^p}<\frac{M}{k^p}\;.$$
Then, as you say, the $p$-test does the rest.