# Convergence of $\sum_{n=1}^{\infty} \sin^{-1}{\frac{1}{n}}$

I am trying to find the convergence of

$$\sum_{n=1}^{\infty} \sin^{-1}{\frac{1}{n}}$$

I tried divergence test but the $\lim = 0$, which is inconclusive? I believe the ratio/root test won't work here too? For comparison / limit comparision test, I think the $\sin^{-1}$ complicates things ... How should I approach this problem?

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Is $\sin^{-1}$ the inverse sine function or $1/\sin(x)$? –  user38268 Mar 18 '12 at 9:55
Please use \sin. –  Did Mar 18 '12 at 10:02
Its inverse $\sin$ –  Jiew Meng Mar 18 '12 at 10:03
@JiewMeng: Isn't it better to use $\arcsin$ instead of $\sin^{-1}$ which can easily be misunderstood as $\frac{1}{\sin}$; at least that's what I thought when I saw the title. –  Beni Bogosel Mar 19 '12 at 9:25

(This answer is just a completion to Didier Piau'a answer). Using the comparison criterion with limit also works here.

Comparison Criterion with Limit Suppose $a_n,b_n>0$ and $\lim_{n \to \infty} \frac{a_n}{b_n}=L \in (0,\infty)$. Then $\sum a_n$ and $\sum b_n$ have the same nature. (i.e. if one converges, so is the other one; if one diverges, so is the other one)

The proof of this criterion is quite simple, and the idea is that for $\varepsilon$ small enough and $n$ large we can write

$$b_n(L-\varepsilon)<a_n<b_n(L+\varepsilon)$$ and $$a_n(\frac{1}{L}-\varepsilon)<b_n<a_n (\frac{1}{L}+\varepsilon)$$

In your case, the fact that $\displaystyle \lim_{n \to \infty} \frac{\arcsin \frac{1}{n}}{\frac{1}{n}}=1 \in (0,\infty)$ simply implies via the criterion above that $$\sum \arcsin \frac{1}{n}$$ has the same nature as $\sum \frac{1}{n}$ which is well known to be divergent.

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Hmm I know that $\lim_{n\to\infty} \frac{\sin{x}}{x} = 1$ but is $\lim_{x\to\infty} \frac{\arcsin{x}}{x} = 1$ a fact too? If so I think $\lim_{x\to\infty} \frac{\tan{x}}{x}=1$ so is $\lim_{x\to\infty} \frac{\tan^{-1}{x}}{x} = 1$ too? –  Jiew Meng Mar 18 '12 at 12:14
Ops ... $x$ should be $\frac{1}{x}$ in my prev comment –  Jiew Meng Mar 18 '12 at 12:17
You didn't write the limits correctly. It is not $x \to \infty$, but $x \to 0$! Yes, theyre all true, and all come from the fact that $\lim_{x \to 0} \frac{\sin x}{x}=1$. –  Beni Bogosel Mar 18 '12 at 12:18
Oh I graphed it and it appears so :) –  Jiew Meng Mar 18 '12 at 12:25

The series diverges.

Since $\sin(x)\sim x$ when $x\to0$, for every $n$ large enough, $\arcsin(1/n)\geqslant1/(2n)$. The series with positive terms $\sum\limits_n1/n$ diverges, hence the series $\sum\limits_n\arcsin(1/n)$ diverges.

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How did $\frac{1}{2n}$ come into the picture? –  Jiew Meng Mar 18 '12 at 10:16
How would you translate $\sin(x)\sim x$ into an inequality valid for every positive $x$ close enough to $0$? –  Did Mar 18 '12 at 10:39
Something like $-1 \le \sin(x) \le 1$? When $x \rightarrow 0$, $\sin(x) \rightarrow 0$ –  Jiew Meng Mar 18 '12 at 10:45
No. Please check the first definition here. –  Did Mar 18 '12 at 11:07