Finding convergence/ divergence of a series- Calculus three

Question

summation of (1/(n)^1/2)-(1/(n+1)^1/2)

 

Does this converge or diverge? I would assume it diverges but online sources say it converges with the comparison test. I don't see how that would work. 

 

Answer 1

Two approaches: telescoping sum, or comparison test with the series ∑ n^-3/2

 

1. Telescoping sum

 

we have ∑(1/(n)^{1/2})-(1/(n+1)^{1/2})

          = lim_N→∞ ∑(1/(n)^{1/2})-(1/(n+1)^{1/2}) , where the sum goes from n=1..N

          = lim_N→∞ [ (1/√1 - 1/√2 + 1/√2 - 1/√3 + ... + 1/√n - 1/√(n+1) + 1/√(n+1) - 1/√(n+2) + ... - 1/√(N+1) ]

          = lim_N→∞ [ 1 - 1/√(N+1) ]

          = 1, converges

 

because the two middle terms in between the "..." are cancelling throughout.

 

2. Comparison test

 

combine fractions, then rationalize the numerator to get

 

1/[√(n²+n)(√(n+1) + √n] < 1/[n√n] = n^-3/2, which is a power series n^-a with a > 1, hence converges