I am doing an exercise but in the middle I am stuck. $$\sum_{k=2}^\infty \frac{[\ln(k+1)-\ln k]^a}{\ln^2k}$$ I have rewritten this as: $$\sum_{k=2}^\infty \frac{[\ln(k+1)/k)]^a}{\ln^2k}$$ Now $a_k=\frac{\ln^a(k+1)/k)}{\ln^2k}$, and apply a Taylor series expansion. Now I wrote: $a_k = \frac{1}{\ln^2k}\left[\frac{1}{k}-\frac{1}{2k^2} + O\left( \frac 1 {3k^3}\right)\right]^a$. By dividing by $\frac{1}{k}$, and applying the binomial expansion, obtaining
$$a_k=\frac{1}{k^a \ln^2k}(1-\frac{a}{2k}+\cdots)$$
But how to proceed? Perhaps I started completely wrong?
HINT:
Using the inequalities
$$\frac{x}{x+1}\le \log(1+x)\le x$$
we see that
$$\frac{1}{(k+1)^a\log^2(k)} \le \frac{\log^a\left(1+\frac1k\right) }{\log^2(k)}\le \frac{1}{k^a\log^2(k)}$$
Can you conclude now?