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I would like to study the convergence of the numerical serie $$ S_n=\sum_{k= 1}^n u_k=\sum_{k= 1}^n \frac{1}{\left(\sqrt[k]{2}+\log k\right)^{k^2}}. $$ I tried the Cauchy rule (i.e. evaluate $\lim_{k\rightarrow +\infty}(u_k)^{\frac 1 k}$ but there is no issue.

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  • $\begingroup$ Question edited. $\endgroup$
    – SAKLY
    Feb 24, 2015 at 17:49

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$$\frac1{\left(\sqrt[k]2+\log k)\right)^{k^2}}<\frac1{\sqrt[k]{2}^{k^2}}$$

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The series $$S=\sum_{k\geq 1}\frac{1}{\left(\sqrt[k]{2}+\log k\right)^{k^2}}$$ is convergent by comparison with a geometric series, since: $$ \left(\sqrt[k]{2}+\log k\right)^{k^2} \geq 2^k,$$ hence $S\leq 1$.

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