Alternating Series Convergence (Conditional and absolute) $$\sum^{\infty}_{k=2}\frac{(-1)^{k+1}}{\sqrt{k(k^2-2)}}$$ Using Leibniz's alternating series test, I determined it's convergent. However when trying to determine absolute convergence, I'm trying to compare it with $\frac{1}{k^{3/2}}$ but I realise that the actual summand is greater than the comparison as denominator is smaller, hence the comparison is invalid to deduce absolute convergence. But answer book says absolute convergence. Please advise me on what I'm doing wrong.

Hint. Set $\displaystyle u_k=\frac{(-1)^{k+1}}{\sqrt{k(k^2-2)}}$, $k\geq 2$.
Then you have
$$
|u_k|=\frac{1}{\sqrt{k(k^2-2)}}=\frac{1}{k^{3/2}\sqrt{1-\dfrac{2}{k^2}}}\sim \frac{1}{k^{3/2}},\quad k \rightarrow +\infty,
$$
and your series $\displaystyle \sum_{k=3}^{\infty}u_k$ is absolutely convergent by the comparison test.
