# Prove that $\frac{1}{\sqrt{2n-1}}-\frac{1}{2n}\geq \frac{1}{2n}$ for $n = 1, 2, 3,…$

Prove that $\frac{1}{\sqrt{2n-1}}-\frac{1}{2n}\geq \frac{1}{2n}$ for $n = 1, 2, 3,...$

This is required to prove that the series $1-\frac{1}{2}+\frac{1}{\sqrt{3}}-\frac{1}{4}$ is divergent, but I don't know how to show this inequality to be true because the square root is present.

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You have an equality, and so you can perform any typical operations you want (such as adding terms to both sides, squaring, multiplying, etc). Try simplifying your equation so that it becomes more clear that it's true, and then you can construct an argument showing that it is.

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just as a quick hint (this was something that I was often confused about while learning these things): if you decrease the denominator you increase the overall fraction – DanZimm Apr 17 '13 at 16:09
Yeah, that's a great point :) – random_forest_fanatic Apr 17 '13 at 16:41

The general strategy with this kind of inequality is to just rearrange it until it is obvious. $\frac{1}{\sqrt{2n-1}} - \frac1{2n} \ge \frac1{2n}$ is equivalent to $\frac{1}{\sqrt{2n-1}} \ge \frac1n$. You can then rearrange further just like you would solve an equation, keeping track of what this does to the $\ge$ sign.

EDIT: I said before it was unclear what series you were talking about, but I was being silly. Of course the series you mean is $\displaystyle \sum_{i = 1}^\infty \left( \frac1{\sqrt{2i - 1}} - \frac1{2i} \right).$

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Hint: Use the comparison test.

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I'm guessing he's trying to show its divergent without using any of the series tests, but rather by showing the limit of the sequence of partial sums does not converge – DanZimm Apr 17 '13 at 16:07