Prove $\sum_{n=1}^\infty \arcsin({\frac 1 {\sqrt{n}}})$ is divergent.

Question

Evaluate if the following series is convergent or divergent: $\sum\limits_{n=1}^\infty \arcsin({\frac 1 {\sqrt{n}}})$.

I think I could apply the integral test that would take me to a complex integral computation. I have checked the solution and the series are divergent. I am trying to employ Weierstrass comparison theorem via the maximization of the function.

Question: How could I prove the series diverge using another test?

Thanks in advance!

Answer 1

Since we have $$\sin x \le x$$

We get $$\arcsin x \ge x$$ therefore, $$ \arcsin \left( \frac{1}{\sqrt n} \right) \ge \frac{1}{\sqrt n} \ge \frac 1n $$

Since $\displaystyle \sum_{n=1}^{\infty} \frac 1n $ diverges, given sum diverges too.

Answer 2

Note that

$$ \arcsin \left( \frac{1}{\sqrt n} \right) \sim \frac{1}{\sqrt n}$$

then the given series diverges by limit comparison test with $\sum \frac{1}{\sqrt n}$.

Answer 3

Since $\arcsin'0=1$ and $\arcsin''x>0,$ for $x>0,$ you get $\arcsin x>x$ for $x>0,$ and thus $\arcsin\dfrac 1 {\sqrt n} > \dfrac 1 {\sqrt{n}}.$ Thus $$ \sum_{n=1}^\infty \arcsin\frac 1 {\sqrt n} \ge \sum_{n=1}^\infty \frac 1 {\sqrt n} = +\infty. $$

What first alerts you to all this is of course that you remember what the graph of the arcsine function looks like.

Answer 4

Another way.-Since $\arcsin(x)=x+\dfrac{x^3}{6}+\dfrac{3x^5}{40}+O(x^7)$ and $0\le \dfrac {1}{\sqrt n}\le 1$ do you have $$\sum_{n=1}^\infty \arcsin\frac 1 {\sqrt n}\gt\sum_{n=1}^\infty(\dfrac {1}{\sqrt n})\gt\sum_{n=1}^\infty \dfrac 1n\to \infty $$

Answer 5

More generally, $\sum\limits_{n=1}^\infty \arcsin({\frac1 {n^a}}) $ converges for $a > 1$ and diverges for $0 < a \le 1$.

This is because $\frac{2}{\pi}x \le \sin(x) \le x$ for $0 \le x \le \frac{\pi}{2}$. Therefore $x \le \arcsin(x) \le \frac{\pi}{2} x $ for $0 \le x \le 1$.

Therefore $\sum\limits_{n=1}^\infty \arcsin({\frac1 {n^a}}) $ converges if and only if $\sum\limits_{n=1}^\infty {\frac1 {n^a}} $ converges.