0
$\begingroup$

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?

  • 0
    @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.2012-03-19

2 Answers 2

1

(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) and $ 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.

  • 0
    Oh I graphed it and it appears so :)2012-03-18
2

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.

  • 0
    No. Please check the first definition [here](http://en.wikipedia.org/wiki/Asymptotic_analysis#Definition).2012-03-18