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

Question

8545 views around the world

You can reuse this answer
Creative Commons License

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#