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If $\sum_{n=2}^{\infty}a_n$ converge, then also converge this series? $$\sum_{n=2}^{\infty}\frac{\sqrt{a_{n}}}{\ln\, n}(n^{a_{n}}-1)$$

Please verify my answer below

Counterexample:

$$a_{n}=\begin{cases} \frac{1}{k^{2}} & n=k!^{k^{2}}\\ \\ 0 & \text{All other cases} \end{cases}$$

When our infinites sum is equal to

$$\sum_{k=1}^{\infty}\frac{1}{k}\cdot\frac{1}{2\, \ln\, k}\cdot(k!-1)$$

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OK, solution moved to the question –  Steve Jan 9 '13 at 19:26
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2 Answers

up vote 1 down vote accepted

The denominator $2\log k$ is supposed to be $\ln n$ but if $n=(k!)^{k^2}$, then $\ln n=k^2\ln(k!)$. Since this is equivalent to $k^3\ln k$, the counterexample holds nevertheless.

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Yes, your counter-example is correct, so your solution is correct.

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This should be a comment as opposed to an answer. –  Amzoti Jan 10 '13 at 0:04
    
Why? It is an answer after all. –  Douglas S. Stones Jan 10 '13 at 0:31
    
Both true and false. –  Did Jan 10 '13 at 0:39
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