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$\frac{1}{1+\sqrt k (-1)^k}= \frac{1-(-1)^k \sqrt k}{1-k}$. Not quite sure how to move on from here.

Labbsserts
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2 Answers2

2

Apply Leibniz test to show that $-\sum\frac {(-1)^{k} \sqrt k} {k-1}$ is convergent. Show that $\sum \frac 1 {1-k}$ is divergent. It then follows that the sum of these two series is divergent.

[$\frac {\sqrt x} {x-1}$ is a decreasing function on $(0,\infty)$ as you can see by differentiation].

Kavi Rama Murthy
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  • For the last paragraph: $\frac{\sqrt x}{x-1}$ is the average of $\frac{\sqrt x-1}{x-1}=\frac1{\sqrt x+1}$ and $\frac{\sqrt x+1}{x-1}=\frac1{\sqrt x-1}$, which are both obviously decreasing on $(1,\infty)$ – Hagen von Eitzen Nov 19 '20 at 06:10
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Rewrite

\begin{align}\sum_{k=2}^{2N+1}\frac{1}{1+(-1)^k \sqrt k}&= \sum_{k=1}^N\left(\frac{1}{1+(-1)^{2k} \sqrt {2k}}+\frac{1}{1+(-1)^{2k+1} \sqrt {2k+1}}\right)\\&=\sum_{k=1}^N\left(\frac{1}{1+\sqrt {2k}}+\frac{1}{1- \sqrt {2k+1}}\right)\\&=\sum_{k=1}^N\left(\frac{1- \sqrt {2k+1}+1+\sqrt {2k}}{(1+\sqrt {2k})(1- \sqrt {2k+1})}\right) \\&<\sum_{k=1}^N\left(\frac{2}{(1+\sqrt {2k+1})(1- \sqrt {2k+1})}\right) \\&=\sum_{k=1}^N\left(\frac{2}{-2k}\right)\to-\infty \end{align}

player3236
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