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

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?

• Would $\sum\limits_{n=1}^\infty \arctan({\frac 1 {\sqrt{n-1}}})$ be easier? First term $=\pi/2.$ – Narasimham Mar 31 '18 at 18:43

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.

Note that

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

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

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.

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.

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 $$\sum_{n=1}^\infty \arcsin\frac 1 {\sqrt n}\gt\sum_{n=1}^\infty(\dfrac {1}{\sqrt n})\gt\sum_{n=1}^\infty \dfrac 1n\to \infty$$

• King Tut: try to factorize $x^7$ after the third addend, of course, and answer your question yourself. Regards. – Piquito Mar 31 '18 at 15:51
• Such an expansion cannot yield this inequality for every $n$, only for $n$ large enough. – Did Apr 2 '18 at 20:27