# Find the value of : $\lim_{n\to\infty}\sqrt[n]{\frac{|\sin1|}1+\cdots+\frac{|\sin n|}{n}\ }$

I just read this question, about a limit very similar to that I am asking. I was confused because I was misreading the product dots in that question as plus signs. The provided, excellent answers are easy to follow, and in fact they allow me to realize about my mistake. Now I am curious about the limit

$$\lim_{n\to\infty}\sqrt[n]{\frac{|\sin1|}1+\cdots+\frac{|\sin n|}{n}\ }\,.$$

I did not try anything, sorry, my only intuition is that the inner sum probably diverges, so its $n$-th root has indeterminate behavior

Clearly the inner sum is bounded between 1 (for $n\geq 2$) and $n$ , and so the limit of the $n$th root is 1 by the squeeze theorem.

The version of this question without absolute signs could be more interesting.

Edit - As pointed out by user 17762, without absolute signs it converges to a constant.

• Without the absolute sign, the limit is $1$ again, since $\displaystyle \sum_{k=1}^{\infty} \dfrac{\sin(k)}k$ is $\dfrac{\pi}2$. – user17762 Dec 7 '13 at 17:21
• @user17762: can you show that or point to something showing it? The sum of the first 1,000 terms is about $1.07$ as is the sum of the first 10,000,000 terms. – Henry Dec 7 '13 at 18:25
• @Henry Well I missed a $-1/2$. We have $\sum \dfrac{\sin(n)}n = \text{Imag}\sum \dfrac{e^{in}}n = -\text{Imag}\log(1-e^i) = \dfrac{\pi-1}2$ – user17762 Dec 7 '13 at 18:35
• @user17762: Thank you - I am convinced – Henry Dec 7 '13 at 18:44
• @user17762 Thanks! – Calvin Lin Dec 7 '13 at 20:24

Upper bound (since $|\sin k| \leq 1$: $$L \leq \lim_{n \to \infty} \bigg( \sum_{k=1}^{n} \frac{1}{k} \bigg)^\frac{1}{n} \sim\lim_{n \to \infty} (\log n)^\frac{1}{n} = \lim_{n \to \infty}e^{\frac{\log \log n}{n}}=1$$ Lower bound: $$L \geq \lim_{n \to \infty}\bigg(\frac{1}{n}\bigg)^\frac{1}{n}=\lim_{n \to \infty} e^{-\frac{\log n}{n}}=1$$

Now use the squeeze lemma