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$$\sum^{\infty}_{k=2}\frac{(-1)^{k+1}}{\sqrt{k(k^2-2)}}$$ 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.

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One way to do it: if $k\ge2$ then $$\frac{k^2}{2}\ge2$$ so $$k^2-2\ge k^2-\frac{k^2}{2}=\frac{k^2}{2}$$ $$\frac{1}{k\sqrt{k^2-2}}\le\frac{1}{k\sqrt{k^2/2}}=\sqrt2k^{-3/2}\ .$$ Now $\sum k^{-3/2}$ converges, $\sum\sqrt2k^{-3/2}$ is a constant times this so also converges,so your series converges by the comparison test.

The limit form of the comparison test is easier, if you have learnt this.

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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.

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Since $k(k^2-2) \ge \dfrac{k^3}2$ for $k\ge 2$, we can apply the comparison test because $$ \frac1{\sqrt{k(k^2-2)}} \le \frac{\sqrt2}{\sqrt{k^3}} = \frac{\sqrt2}{k^{3/2}} $$ assuming that you know that $\sum \dfrac1{k^{\alpha}}$ converges for all $\alpha>1$.

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You can use the substitution $u=k-2$, so the absolute value of the sum is equal to $$\sum\limits_{u=0}^\infty \frac{1}{\sqrt{(u+2)(u^2+4u+2)}} < \sum\limits_{u=0}^\infty \frac{1}{\sqrt{u^3}}$$ because $(u+2)(u^2+4u+2) > u^3$ for all natural numbers $u$.

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  • $\begingroup$ I suggest letting $u=k-1$ because now you have division by zero on the right. $\endgroup$ – Adayah Nov 4 '14 at 19:21

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