This was a recent problem on the Awesome Math Problem Column. The solution is given as follows:
We shall use Stolz-Cesaro Lemma. We have:
$$\lim_{n\to\infty} \frac{\sin(1)+\sin^2(\frac{1}{2})+\ldots+\sin^n(\frac{1}{n})}{\frac{1}{1!}+\frac{1}{2!}+\ldots+\frac{1}{n!}}$$ $$=\lim_{n\to\infty}\frac{\sin^n(\frac{1}{n})}{\frac{1}{n!}}=\lim_{n\to\infty}\frac{\sin^n(\frac{1}{n})}{\frac{1}{n^n}}\cdot\lim_{n\to\infty} \frac{n!}{n^n}.$$ Since $$\frac{n!}{n^n}<\frac{1}{n}$$ we have $$\lim_{n\to\infty}\frac{n!}{n^n}=0$$ So, the required limit is $0$.
I have two problems with the given proof that I can’t wrap my head around.
Firstly, from everything I have read so far the Stolz-Cesaro limit only applies to cases where the limit is of the form $\frac{0}{0}$ and the form $\frac{\cdot}{\infty}$. However, just from the wording of the theorem and the way the author presented this solution both seem to imply the following: $$a_n=\sum_{k=1}^n \sin^k\Big(\frac{1}{k}\Big), \ \ \ \text{and} \ \ \ b_n=\sum_{k=1}^n \frac{1}{k!}$$ Then when “applying” the Stolz-Cesaro theorem you would have: $$\lim_{n\to\infty} \frac{a_n-a_{n-1}}{b_n-b_{n-1}}=\lim_{n\to\infty} \frac{\sum\limits_{k=1}^{n} \sin^k(\frac{1}{k})-\sum\limits_{k=1}^{n-1} \sin^k(\frac{1}{k})}{\sum\limits_{k=1}^n \frac{1}{k!}-\sum\limits_{k=1}^{n-1} \frac{1}{k!}}=\lim_{n\to\infty}\frac{\sin^n(\frac{1}{n})}{\frac{1}{n!}}$$
The problem that I see is when $n\to\infty$ we know that $a_n$ converges by the root test, and $b_n$ converges to $e-1$. So by basic limit rules this would converge also. So it seems that $b_n$ does not satisfy either of the sequence requirements of the Stolz-Cesaro Theorem.
Secondly, according to wolfram $\displaystyle\sum_{n=1}^\infty \sin^n\Big(\frac{1}{n}\Big)\approx 1.11043$. So intuitively it seems to me that this limit would be approximately $\frac{1.11043}{e-1}\approx 0.646244$.
Does anyone know what is going on here?
Note sure that this is a valide answer.
Let $$a_k=\sin ^k\left(\frac{1}{k}\right)\implies \log(a_k)=k \log \left(\sin \left(\frac{1}{k}\right)\right)$$ Using Taylor $$\log(a_k)=-k \log (k)-\frac{1}{6 k}-\frac{1}{180 k^3}+O\left(\frac{1}{k^5}\right)$$ $$a_k=k^{-k}\,\exp\left(-\frac{1}{6 k}-\frac{1}{180 k^3}+\cdots \right)$$ The summation of the $a_k$'s from $k=1$ to $n$ converges very fast to $1.110750$ which, divided by $e-1$ gives $0.646431$ while the result should be $0.646242$. Adding the next term in the exponential (namely $-\frac{1}{2835 k^6}$) would give $0.646256$ and so forth since all terms in the exponential are negative.
Notice that $$\sum_{k=1}^\infty k^{-k}=1.2912859970626635404\cdots$$ is a well know number (have a look here). Forgetting the exponential term, this would give an upper bound equal to $\frac 34$ for the limit.