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$$\sum_{k=1}^{\infty}\left(\frac{1}{k^{\log k}}\right)$$ this is converge. But how? I think here we use comparison thm.

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Please avoid using $$ and \displaystyle in the title. – Asaf Karagila Dec 27 '12 at 21:02

Try comparing this series to $$\sum_{k=1}^\infty \frac{1}{k^2}.$$ Since $\log(x)\geq 2$ for $x\geq e^2$, after throwing out finitely many terms you can prove convergence.

Added: Alternatively, by using the Cauchy condensation test, we need only prove the convergence of $$\sum_{n=0}^{\infty}\frac{2^n}{2^{n^{2}\log2}}.$$

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Let $a_k = 1/k^{\log k}$. Then $$\begin{eqnarray*} \frac{a_{k+1}}{a_k} &=& \frac{k^{\log k}}{(k+1)^{\log (k+1)}} \\ &=& \exp \left[\log^2 k - \log^2(k+1)\right]. \end{eqnarray*}$$ Now show $\log^2 k < \log^2(k+1)$ for $k\ge 1$ and apply the ratio test.

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