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.

Let $0 < a < 1$. I'm trying to figure out whether the following series converges:

$$\sum_{k=1}^\infty k^a \frac{1}{k(k+1)}.$$

Now it's clear that if $a$ were greater than or equal to $1$ then this series would diverge since

$$\sum_{k=1}^\infty k^1 \frac{1}{k(k+1)} = \sum_{k=1}^\infty \frac{1}{(k+1)} = \sum_{k=2}^\infty \frac{1}{k} = \infty.$$

So this makes it a bit hard to think of a bound for the series in question. Any advice?

share|improve this question

1 Answer 1

up vote 4 down vote accepted

$$0\leq \frac{k^\alpha}{k(k+1)}\leq \frac 1{k^{2-\alpha}},$$ and $2-\alpha>1$.

share|improve this answer
    
In addition to Davide's answer it may be noted that $\frac{1}{2k^{2-\alpha}}\leqslant \frac{k^\alpha}{k(k+1)}\leqslant\frac 1{k^{2-\alpha}},$ so given series diverges for $2-\alpha \leqslant{1}.$ –  M. Strochyk Nov 1 '12 at 16:34

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.