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Fix a positive number $\alpha$ and consider the series $$\sum_{k=1}^\infty \frac1{(k+1)[\ln(k+1)]^\alpha} $$

For what values of $\alpha$ does this series converge?

I can plug in a bunch of values, but how would you show this mathematically?

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Do you know the Integral Test? –  André Nicolas Dec 2 '11 at 7:28
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2 Answers

up vote 2 down vote accepted

Consider the function $\displaystyle\frac{1}{(x+1)[ln(x+1)]^\alpha}$ for $\alpha>0$. Then $\displaystyle\frac{1}{(k+1)[ln(k+1)]^\alpha}$ is positive, decreasing, continuous on $[2,\infty)$. Therefore, by integral test, we have $$\displaystyle\sum_{k=1}^\infty \displaystyle\frac{1}{(k+1)[ln(k+1)]^\alpha}$$ converges if and only if $$\int_1^\infty\frac{1}{(x+1)[ln(x+1)]^\alpha}dx$$ converges. On the other hand, if $\alpha>1$, then $$\int_1^\infty\frac{1}{(x+1)[ln(x+1)]^\alpha}dx=\frac{1}{1-\alpha}\ln(x+1)^{1-\alpha}\Big|^\infty_1<\infty,$$ and it diverges if $0<\alpha\leq 1$. Therefore, the series converges when $\alpha>1$ and diverges when $0<\alpha\leq 1$.

Also, for $\alpha\leq 0$, we have $$\displaystyle\frac{1}{(k+1)[ln(k+1)]^\alpha}\geq\frac{1}{k+1}$$ and the harmonic series $\displaystyle\sum_{k=1}^\infty\frac{1}{k+1}$ diverges. By comparison test, $\displaystyle\sum_{k=1}^\infty \displaystyle\frac{1}{(k+1)[ln(k+1)]^\alpha}$ diverges when $\alpha\leq 0$.

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$\displaystyle \sum_{n=2}^{\infty} \dfrac{1}{n (\log n)^{\alpha}}$ is convergent iff $\displaystyle \alpha > 1$

This can be checked by using the well known test that if we have a monotone decreasing sequence, then $\displaystyle \sum_{n=2}^{\infty} a_n$ converges iff $\displaystyle \sum_{n=2}^{\infty} 2^na_{2^n}$ converges (or) by the integral test.

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