Using the integral test, how do you show whether sum 1 / [sqrt(n) * (sqrt(n) + 1)]1n(n+1) diverges or converges from n=1 to infinity?

1 Answer
Jun 13, 2016

The sum is unbounded

Explanation:

We know

sum_{i=1}^{n} 1 / [sqrt(i) * (sqrt(i) + 1)]geint_{1}^n(dx)/(sqrt(x)(sqrt(x)+1)} = 2log_e(1+sqrt(n))ni=11i(i+1)n1dxx(x+1)=2loge(1+n)

because 1 / [sqrt(i) * (sqrt(i) + 1)]1i(i+1) is monotonic decreasing.

but lim_{n->oo}2log_e(1+sqrt(n)) = oo

so the sum is unbounded

Attached a figure with the comparisson between series and integral.

enter image source here