How do you test the alternating series $Sigma (-1)^n(sqrt(n+1)-sqrtn)$ from n is $[1,oo)$ for convergence?

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Answer 1 · 2017-11-01T12:47:54

See below.

We have

$sqrt(n+1)-sqrt(n) le 1/2 1/sqrtn$ so the series

$sum_(k=1)^oo (-1)^n(sqrt(n+1)-sqrtn) le 1/2 sum_(k=1)^oo(-1)^n 1/sqrt(n)$

but the series with $a_n = (-1)^n1/sqrtn$ is alternating and absolutely convergent then

$sum_(k=1)^oo (-1)^n(sqrt(n+1)-sqrtn)$ is convergent.

NOTE:

From the mean value theorem, with $f(x) in CC^2$ there exists $xi in (a,b)$ such that

$f(b)-f(a) = f'(xi)(b-a)$

Calling $f(x) = sqrtx$ , $a = n$ , $b = n+1$ and knowing that in this case

$f'(x)$ is monotonically decreasing in $n, n+1$ we establish that

$sqrt(n+1)-sqrt(n) le 1/2 1/sqrtn$

How do you test the alternating series Sigma (-1)^n(sqrt(n+1)-sqrtn)Sigma (-1)^n(sqrt(n+1)-sqrtn) from n is [1,oo)[1,oo) for convergence?