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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?

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

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  • $\begingroup$ 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 $2-\alpha \leqslant{1}.$ $\endgroup$ – M. Strochyk Nov 1 '12 at 16:34

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