Application of Leibniz Criterion to $\sum_{k=2}^{+\infty}(k^2+1)\sin(\frac{1}{k^3-1})\alpha^k$

Question

I have to discuss absolute and simple convergence of $\sum_{k=2}^{+\infty}(k^2+1)\sin(\frac{1}{k^3-1})\alpha^k$,where $\alpha\in\mathbb{R}$.
If $\alpha=-1$ I have $\sum_{k=2}^{+\infty}(-1)^kb_k$,where $b_k=(k^2+1)\sin(\frac{1}{k^3-1})$.By Leibniz Criterion,for convergence I must have:
i)$\lim_{k\to+\infty}b_k=0$
ii)$b_k$ eventually decreasing for $k\to+\infty$
For the first condition $\lim_{k\to+\infty}=(k^2+1)\sin(\frac{1}{k^3-1})\sim\lim_{k\to+\infty}\frac{1}{k}=0$
As for the second condition ${\frac{1}{k}}'=-\frac{1}{k^2}\lt0$.Can I deduce from this that our original $b_k$ is also decreasing,or I have to study derivative of $b_k$ itself?

Answer 1

By applying ratio test, we know that the series is absolutely convergent for $|\alpha|<1$ and divergent for $|\alpha|>1$. For $\alpha=1 $ the series is divergent since $$(k^2+1)\sin(\frac{1}{k^3-1})\sim {k^2+1\over k^3-1}\sim {1\over k}$$similarly for $\alpha=-1$$$(k^2+1)\sin(\frac{1}{k^3-1})(-1)^k\sim {(-1)^k\over k}$$which is convergent but not absolutely.