Question

test the convergence of the series sigma tan inverse 1/n

Answer 1

tan(x)=sin(x)cos(x)tan⁡(x)=sin⁡(x)cos⁡(x).

For 0≤x≤π/20≤x≤π/2, sin(x)≤xsin⁡(x)≤x. Integrating this from 00 to tt, ∫t0sin(x)dx≤∫t0xdx∫0tsin⁡(x)dx≤∫0txdx or −cos(x)|t0≤x22|t0−cos⁡(x)|0t≤x22|0t or 1−cos(t)≤t2/21−cos⁡(t)≤t2/2 or cos(t)≥1−t2/2cos⁡(t)≥1−t2/2.

Therefore tan(x)≤x1−x2/2tan⁡(x)≤x1−x2/2 so that, for n≥1n≥1, tan(1/n)≤1/n1−(1/n)2/2<2/ntan⁡(1/n)≤1/n1−(1/n)2/2<2/n so that tan2(1/n)<4/n2tan2⁡(1/n)<4/n2 and the sum of these converges by any test you want to use.

hope it helps you
please add to Brian list