Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.