# Divergence/convergence of the series [closed]

Given that $\frac{1}{\sqrt{x}} \ge \frac{1}{x+1}$ for all $x> 0$, determine the convergence or divergence of the series, $$\sum_{k=1}^{\infty}\left( \frac{1}{\sqrt{2k-1}} - \frac{1}{2k} \right)$$

Any hints would be much appreciated, this is really making my head hurt. I'm sure the comparison test must be involved, but I can't seem to figure out how. Thanks.

## closed as off-topic by Hurkyl, Xander Henderson, Taroccoesbrocco, José Carlos Santos, NamasteAug 22 '18 at 10:59

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• Estimate the term. – xbh Aug 15 '18 at 4:41
• $1/\sqrt {2k-1} > 1/ \sqrt {2k}$, and ... – xbh Aug 15 '18 at 4:48

$$\frac{1}{\sqrt{2k-1}} - \frac{1}{2k}=\frac{2k-\sqrt{2k-1}}{2k\sqrt{2k-1}}\sim\frac{1}{\sqrt{2k-1}}\ \text{as}\ k\to\infty.$$ So by comparison test your series is divergent.

With $(k-1)^2=k^2-2k+1>0$, for $k>1$ then $\sqrt{2k-1}<k$ therefore $$\dfrac{1}{\sqrt{2k-1}}-\dfrac{1}{2k}>\dfrac{1}{2k}$$

I think I have it.

Given that $\frac{1}{\sqrt{x}} \geq \frac{2}{x+1}$, and let $x = 2k - 1$

$$\frac{1}{\sqrt{2k-1}} - \frac{1}{2k} \geq \frac{2}{2k-1+1} - \frac{1}{2k}= \frac{1}{k} - \frac{1}{2k}=\frac{1}{2k}$$

And since,

$$\sum_{k=1}^{\infty}\frac{1}{2k} = \frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{k}$$

which diverges. Hence by the comparison test, $\sum_{k=1}^{\infty}\left( \frac{1}{\sqrt{2k-1}} - \frac{1}{2k} \right)$ also diverges

Does this seem right? Sorry it's pretty basic, we're only allowed to assume anything that we've proven in lectures or problem sets. I appreciate all your answers.