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

1 Answer
Nov 1, 2017

See below.

Explanation:

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