# $\lim_{x\to\infty} \frac{\sin x+\sin^2x+\sin^3x+\dotsb}{x\sin x}$, infinite series limit

I saw this question yesterday.

$$\lim_{x\to\infty} \frac{\sin x+\sin^2x+\sin^3x+\dotsb}{x\sin x}$$

I claim that the limit is $0$ because it can be written like the following.

$$\lim_{x\to\infty} \left(\frac{1}{x}+\frac{\sin x}{x}+\frac{\sin^2x}{x}+\dotsb\right)$$

Then I say in this expression the numerator is limited between $[-1,1]$, and the denominator for each term goes as much as to infinity. Hence term by term we have 0, and adding these up we have 0 as the answer.

But then some guy says its indeterminate, because

$$\lim_{x\to\infty} \frac{1+\sin x+\sin^2x+\dotsb}{x} = \lim_{x\to\infty} \frac{1-\sin^nx}{(1-\sin x)x}$$

He claims what we have in the denominator is oscillating and we cannot have a limit. Also, he says I'm wrong because an infinite amount of zero does not equal to zero.

Let

$$f(x)=\frac{1+\sin x+\sin^2x+\ldots}x$$

for all $x\in\Bbb R$ for which this is defined. If $x>0$ and $\sin x\ne\pm 1$, then

$$f(x)=\frac{1}{x(1-\sin x)}\;.$$

Let $n$ be any odd positive integer. By taking $x$ sufficiently close to $\frac{n\pi}2$, we can make$f(x)$ arbitrarily large. Thus, there is a strictly increasing sequence $\langle x_n:n\in\Bbb N\rangle$ of positive real numbers such that $\lim\limits_{n\to\infty}f(x_n)=\infty$.

Since there is also such a sequence for which $f(x_n)=0$ for each $n\in\Bbb N$, the limit does not exist.

• Is such a kind of method always applicable? Even in exceptional cases? Commented Jul 27, 2022 at 8:31
• @Ritil: I don’t really understand the question, since I don’t see this as being an example of any particular method. Commented Jul 28, 2022 at 2:45

The expression does not have a limit as $x \to \infty$ as pointed out by the comment, but you can figure it out on your own. To see this, if $\sin x = 1 \Rightarrow x = \dfrac{\pi}{2} + 2n\pi \Rightarrow$ the expression equals to $\infty$ for any $n \in \mathbb{N}$, thus by taking the limit as $n \to \infty$ it goes to $\infty$ as well. But for those $x$ such that $\sin x = \dfrac{1}{2}$ that is $x = \dfrac{\pi}{6}+ 2n\pi$, the expression is dominated by $\dfrac{10}{n}$, hence using squeeze theorem it goes to $0$ when $n \to \infty$. This shows there are two different answers, so that means we have no limit.

Not. Indeed we have $\lim_{x\to\infty} \dfrac{1+\sin x+\sin^2x+\cdots}{x} = \lim_{x\to\infty} \dfrac{1}{(1-\sin x)x}$ which doesn't exist (cause $f(x)=\dfrac{1}{(1-\sin x)x}$ tends to $\infty$ for $x=\frac{\pi}{2}+2\pi n$ and is $0$ in $x=2\pi n$).

$$\lim_{x\to\infty}\space\frac{\sin(x)+\sin^2(x)+\sin^3(x)+\dots}{x\sin(x)}=$$ $$\lim_{x\to\infty}\space\frac{\sin^1(x)+\sin^2(x)+\sin^3(x)+\dots}{x\sin(x)}=$$ $$\lim_{x\to\infty}\space\frac{\sum_{k=1}^{\infty}\sin^k(x)}{x\sin(x)}=$$

Notice:

$$\sum_{k=1}^{\infty}\sin^k(x)=-\frac{\sin(x)}{\sin(x)-1}\space\space\space\space\space\text{when}\space\space|\sin(x)|<1$$

$$\lim_{x\to\infty}\space\frac{-\frac{\sin(x)}{\sin(x)-1}}{x\sin(x)}=\lim_{x\to\infty}\space\frac{1}{x(\sin(x)-1)}\to\space\text{this limit does not exist}$$

• This reproduces already posted answers but leaves unexplained the crucial step. -1.
– Did
Commented Jan 17, 2016 at 10:47