Prove that series $\sum_{n=1}^\infty \arcsin(\frac{1}{\sqrt n})$ diverges I know that series $\sum_{n=1}^\infty \arcsin(\frac{1}{\sqrt n})$ diverges using the comparison test: $\arcsin(\frac{1}{\sqrt n}) \ge  \frac{1}{\sqrt n} \ge \frac{1}{n} \ge 0$ for $n=1, 2, ...$ But how can I prove that $\arcsin(\frac{1}{\sqrt n}) \ge  \frac{1}{\sqrt n}$?
I think one can show that $\sin x \le x$, so $\arcsin x  \ge x$ for $x \ge 1$, but how can i do that if I am not allowed to use Taylor series? Basically, I am only allowed to use the facts from Calculus I.
 A: Anyway you can prove it diverges using equivalents: it is a series with positive terms, and $\arcsin\frac1{\sqrt n}\sim_\infty \frac1{\sqrt n}$. The latter diverges, hence the former diverges too.
A: From 
$$
\sqrt{1-x^2} \leq1, \qquad 0< x<1,
$$ you get
$$
\frac1{\sqrt{1-x^2}}\geq 1 , \qquad 0< x<1,
$$ that is
$$
(\arcsin x)' \geq (x)'
$$ then conclude with the observation that $\arcsin 0=0$ and put $x=\frac1{\sqrt{n}}$, $n=1,2,\ldots.$
A: You don't need to compare which of them is larger. Instead, 
I like to use the limit comparison test as below.
\begin{align}
\lim_{n\rightarrow\infty}\frac{\arcsin\left(\frac{1}{\sqrt{n}}\right)}{\frac{1}{\sqrt{n}}}
=\lim_{t\rightarrow0^+}\frac{\arcsin(t)}{t}
\stackrel{{\rm H}}{=}\lim_{t\rightarrow0^+}\frac{\frac{1}{\sqrt{1-t^2}}}{1}=1.
\end{align}
Thus $\sum\frac{1}{\sqrt{n}}$ diverges implies $\sum\arcsin\left(\frac{1}{\sqrt{n}}\right)$ diverges.
A: It depends on the contents of Calculus I. In the standard simultaneously thick and thin North American calculus book, something like the following approach is taken. 
Let $t$ be small positive. Take a circle with centre $O$ and radius $1$. Let $A$ and $B$ be points on the circle with $\angle AOB=2t$. Then the arc $AB$ has length $2t$. 
The length of the chord $AB$ is $2\sin t$. (Drop a perpendicular from $O$ to $M$ on $AB$. Then $AM=\sin t$)
The chord has length less than the  arc. It follows that $2\sin t\lt 2t$ and therefore $\sin t\lt t$. 
A: To show that $\arcsin\left(\frac{1}{\sqrt n}\right)\ge \frac{1}{\sqrt n}$, we recall from elementary geometry that the sine function satisfies the inequalities SEE THIS ANSWER
$$\theta \cos(\theta)\le \sin(\theta)\le \theta \tag 1$$
for $0\le \theta \le \pi/2$. 
Then, substituting $\theta =\arcsin(x)$ in $(1)$ yields
$$\arcsin(x)\sqrt{1-x^2}\le x\le \arcsin(x) \tag 2$$
for $0\le x\le 1$.  Therefore, for $0<x<1$ we have
$$x \le \arcsin(x)\le \frac{x}{\sqrt{1-x^2}} \tag 3$$
Finally, letting $x=\frac1{\sqrt n}$ in $(3)$ we obtain
$$\frac1{\sqrt n}\le \arcsin\left(\frac1{\sqrt n}\right)\le \frac{1}{\sqrt{n-1}}$$
for $n> 1$
Therefore, by the comparison test, the series of interest diverges. 
A: Not sure what they teach you in Calculus 1 (where I grew up this would be junior year in high school), but: First, I believe you mean $x\le 1$; second, does your Calculus 1 cover this:  $f(x) = x - \sin x$ has value $0$ at $ x = 0$ and has derivative $1 - \cos x \ge 0$, so $f$ is non-decreasing, so $f(x) \ge 0$ for all $x \ge 0$?
A: It's enough to show that
$\sin(x)
\le x
$
for
$x \ge 0$,
since this gives
$\arcsin(x)
\ge x
$.
Here's another proof:
Since
$\cos(x)
=\sin'(x)
$,
then,
for
$0 \le x \le \frac{\pi}{2}$,
$\begin{array}\\
\sin(x)
&=\int_0^x \cos(t)dt\\
&\le\int_0^x dt
\qquad\text{since }0\le \cos(t) \le 1\\
&\le x\\
\end{array}
$
