# Can some one help me? sum_(n=1)^∞(sqrt(n+2)-2sqrt(n+1)+sqrt(n))

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#### Explanation

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1
Feb 9, 2018

See below.

#### Explanation:

This is equivalent to

${\sum}_{n = 3}^{\infty} \sqrt{n} + {\sum}_{n = 1}^{\infty} \sqrt{n} - 2 {\sum}_{n = 2}^{\infty} \sqrt{n} = 1 - \sqrt{2}$

Then teach the underlying concepts
Don't copy without citing sources
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#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Max G. Share
Feb 9, 2018

Telescoping Series 1

#### Explanation:

$\Sigma \left(\sqrt{n + 2} - 2 \sqrt{n + 1} + \sqrt{n}\right)$
$\Sigma \left(\sqrt{n + 2} - \sqrt{n + 1} - \sqrt{n + 1} + \sqrt{n}\right)$
$\Sigma \left(\left(\sqrt{n + 2} - \sqrt{n + 1}\right) \left(\frac{\sqrt{n + 2} + \sqrt{n + 1}}{\sqrt{n + 2} + \sqrt{n + 1}}\right) + \left(- \sqrt{n + 1} + \sqrt{n}\right) \left(\frac{\sqrt{n + 1} + \sqrt{n}}{\sqrt{n + 1} + \sqrt{n}}\right)\right)$
Sigma(1/(sqrt(n+2) + sqrt(n+1))+(-1)/(sqrt(n+1) + sqrt(n))))
This is a collapsing (telescoping) series.
Its first term is
$- \frac{1}{\sqrt{2} + 1} = 1 - \sqrt{2}$.

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