Series Convergence for a variable 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?
 A: 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  $\alpha\leq 0$. 
A: $\displaystyle \sum_{n=2}^{\infty} \dfrac{1}{n (\log n)^{\alpha}}$ 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.
