divergence of $ \sum_{n=1}^{\infty}\arcsin(1/n)$

Question

$ \sum_{n=1}^{\infty}\arcsin(1/n)$ need some hint to prove that this sum is DIVERGENT

i have tried wolfram alpha and the partial series seems to go increasing so $ S(k)<S(k+1) $

but no hint after there to prove convergence

One way is to use integral test (because all terms are positive): $\int \sin^{-1}(1/x) dx$ is elementary. Solution involves $u$-sub of $u = 1/x$ followed by integration by parts. — Doug, Aug 12 '22 at 17:40

score 3 · Accepted Answer · answered Aug 12 '22 at 17:54

3

$\arcsin \frac 1n \ge \frac 1n$ and $\sum \frac 1n$ diverges.

answered Aug 12 '22 at 17:54

Kroki

13,135

@SouravGhosh thanks for the +1. I think it is a complete answer. – Kroki Aug 12 '22 at 18:26

score 2 · Answer 2 · edited Aug 12 '22 at 17:56

Solution 1 (more difficult): Use integral test (because all terms are positive): $\int \sin^{-1}(1/x)dx$ is elementary. Solution involves u-sub of u=1/x followed by integration by parts.

Solution 2 (easier): Because $\sin^{-1}$ is monotonic on the interval $(0,1]$, and $\sin(x)<x$, it follows that $x<\sin^{-1}(x)$. Therefore, your series is larger than the harmonic series.

In either case, you will find that the series diverges.

divergence of $ \sum_{n=1}^{\infty}\arcsin(1/n)$

2 Answers2