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How to prove that $\sum \frac {1}{(n+3)\ln^ 3 (n+3)} $ converges?

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Just as David pointed out:

$$\int\limits_1^\infty\frac1{(x+3)\log^3(x+3)}dx=\left.-\frac12\frac1{\log^2(x+3)}\right|_1^\infty=\frac12\frac1{\log^24}=\frac18\frac1{\log^22}$$

and thus our series is convergente, or by the Condensation Test:

$$\frac{2^n}{(2^n+3)\log^3(2^n+3)}\le\frac1{n^3\log^32}$$

and since the series with general term $\;n^{-3}\;$ converges so does ours.

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  • $\begingroup$ I think as in the title, the OP wanted the integral version. +1 $\endgroup$ – mrs Dec 24 '13 at 18:08
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The Bertrand series $$\sum_{n\ge2}\frac{1}{n^\alpha\ln^\beta n}$$ is convergent if and only if $(\alpha>1)\lor(\alpha=1\land\beta>1).$

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