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I am asked to study the convergence of the following series: $$\sum_{n=1}^\infty \frac{1}{n(1+ln(n))^\alpha}$$

I have been told that the answer is that the aforementioned series is convergent when alpha is bigger than 1, yet I am at a loss at how to prove it.

I suppose I should apply the limit comparison test, but I do not know which other succession I should pick.

Thanks in advance!

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HINT: $\sum a_n$ converge iff $\sum 2^n a(2^n)$ converge (Known as the Cauchy Condensation Test)

So the problem is reduced to check the convergence of $$\sum \frac{2^n}{2^n(1+n\ln 2)^\alpha}=\sum \frac{1}{(1+n\ln 2)^\alpha}$$ Note $$\sum \frac{1}{(\ln 2+n\ln 2)^\alpha}\le\sum \frac{1}{(1+n\ln 2)^\alpha}\le\sum \frac{1}{(1+n)^\alpha}$$ $$\Rightarrow \frac{1}{(\ln 2)^\alpha}\sum \frac{1}{(1+n)^\alpha}\le\sum \frac{1}{(1+n\ln 2)^\alpha}\le\sum \frac{1}{(1+n)^\alpha}$$

$\sum 1/(1+n)^\alpha$ is the well-known $p$-series which converge if $\alpha >1$.(Not $\alpha \le 1$ as stated)

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Comparing it to an analogous integral, $$\int_1^\infty \frac{dx}{x(1+\ln{x})^\alpha}$$ and with a substitution $u=\ln{x}$, you obtain $$\int_0^\infty \frac{du}{(1+u)^\alpha}$$ which converges for $\alpha>1$ (not less than 1 as the problem states).

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  • $\begingroup$ Right, my bad I messed up copying the correct answer at class! $\endgroup$ – NFC Nov 9 '16 at 13:34

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