So far I got: 2 n + 1 &gt; 2 n − 1 , so 2 n + 1 − 2 n − 1 &gt; 0 for all n &gt; 0

Question: "So far I got: 2 n + 1..."

Emery Boone

2022-05-22

So far I got:

\sqrt{2 n + 1} > \sqrt{2 n - 1}

, so

\sqrt{2 n + 1} - \sqrt{2 n - 1} > 0

for all

n > 0

a2g1g9x

Beginner2022-05-23Added 12 answers

The trick when you have this type of expression is to get rid of the square roots. You can use

a^{2} - b^{2} = (a - b) (a + b)

In this case you have

a = \sqrt{2 n + 1}

and

b = \sqrt{2 n - 1}

, so

\sqrt{2 n + 1} - \sqrt{2 n - 1} = \frac{(\sqrt{2 n + 1})^{2} - (\sqrt{2 n - 1})^{2}}{\sqrt{2 n + 1} + \sqrt{2 n - 1}} = \frac{2}{\sqrt{2 n + 1} + \sqrt{2 n - 1}}

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