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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.

HeroZhang001
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Vika
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  • You just use the comparison test. No need to use Taylor series, you literally just write $x=\arcsin(sin x) $ (this is true for small x, i.e. $x<1$, so your second inequality is literally the same inequality as your first one. – Adam Hughes Mar 26 '16 at 21:19
  • How can one prove that "this is true for small x, i.e. x<1"? – Vika Mar 26 '16 at 21:42
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    It follows from the definition of arcsine. No need to prove, but do be explicit. –  Mar 26 '16 at 21:53
  • @Vika because arcsin takes values in $-\pi/2\le x\le \pi/2$ – Adam Hughes Mar 27 '16 at 03:18

7 Answers7

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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.

Bernard
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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 $x=\frac1{\sqrt{n}}$, $n=1,2,\ldots.$

Olivier Oloa
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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.

Solumilkyu
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  • This works. Thank you. – Vika Mar 26 '16 at 21:43
  • @Vika This is the content of Bernard's answer, but expressed less obscurely. – Michael E2 Mar 26 '16 at 21:45
  • @MichaelE2 I fail to spot what would be "obscure" in the other answer. – Did Mar 26 '16 at 21:46
  • @MichaelE2 An answer is "obscure" because it uses the sign $\sim$ to denote equivalent sequences? Come on. – Did Mar 26 '16 at 21:52
  • @Did (1) I compared the clarity and did not say it was obscure. (2) While common notation & terminology may vary by country, someone asking a basic intro calc. question in the US is unlikely to understand Bernard's terminology and notation. I was attempting to point out to Vika the meaning of Bernard's answer. (Admittedly, it was presumptuous to assume Vika did not understand, but I saw Vika's comment here before the other.) (3) I was also pointing out to Solumilkyu that this answer duplicates Bernard's earlier answer, although this answer has more complete details. -- What's the big deal? – Michael E2 Mar 26 '16 at 21:58
  • The (very small) deal here is that "obscurely" is unwarranted, as mentioned in my very first comment. You seem to more or less withdraw the term now, which is a good thing (which could have been done more simply). – Did Mar 27 '16 at 08:23
  • I think you misread my comment. It was meant as a criticism of this answer, not Bernard's, whose priority I thought I was defending. In hindsight, I might have said "more clearly" instead of "less obscurely" and not triggered your response. I still don't know whether $\sim_\infty$ means asymptotic at infinity (my first reading); or that the ratio approaches a positive limit, or that the ratio is bounded by positive numbers in a neighborhood of infinity (possible meanings of equivalent). – Michael E2 Mar 27 '16 at 15:06
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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 $\sin t\lt t$.

André Nicolas
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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.

Community
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Mark Viola
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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 $x \ge 0$?

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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} $

marty cohen
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