How to find $ \lim_{n\to\infty}\frac{\sin(1)+\sin(\frac{1}{2})+\dots+\sin(\frac{1}{n})}{\ln(n)}$ 
$$ \lim_{n\to\infty}\frac{\sin(1)+\sin(\frac{1}{2})+\dots+\sin(\frac{1}{n})}{\ln(n)}
$$

I tried applying Cesaro Stolz and found its $(\sin 1/(n+1))/\ln(n+1)/n$ where $\ln$ is $\log_e$ and it would be $1$ and so the limit is $0$, but in my book the answer is $2$. Am I doing something wrong or can't Cesaro be applied here?
 A: $$\ \lim_{n\to+\infty}\frac{\sin(1)+\sin(\frac{1}{2})+\dots+\sin(\frac{1}{n})}{\ln(n)}=\lim_{n\to+\infty}\frac{\sum_{k=1}^n\sin(\frac{1}{k})}{\ln(n)}$$
Applying Stolz-Cesàro yields
$$\ \lim_{n\to+\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\lim_{n\to+\infty}\frac{\sum_{k=1}^{n+1}\sin(\frac{1}{k})-\sum_{k=1}^n\sin(\frac{1}{k})}{\ln(n+1)-\ln(n)}= \lim_{n\to+\infty}\frac{\sin(\frac{1}{n+1})}{\ln(n+1)-\ln(n)}=$$
$$\ =\lim_{n\to+\infty}\frac{\sin(\frac{1}{n+1})}{\ln(1+\frac{1}{n})}\sim \lim_{n\to+\infty}\frac{\frac{1}{n+1}}{\frac{1}{n}}=\lim_{n\to+\infty}\frac{n}{n+1}=1$$
Where I used the fact that for $\ x\to 0$, $\sin(x)\sim x$, and $\ln(1+x)\sim x$
A: If your aim is to compute
$$
\lim_{n\to\infty}\frac{\sin1+\sin\frac{1}{2}+\dots+\sin\frac{1}{n}}{\ln n}
$$
you can indeed try and apply Stolz-Cesàro with
$$
a_n=\sin1+\sin\frac{1}{2}+\dots+\sin\frac{1}{n}
\qquad
b_n=\ln n
$$
This leads to computing
$$
\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}
=
\lim_{n\to\infty}\frac{\sin\frac{1}{n+1}}{\ln(n+1)-\ln n}
$$
This limit will exist if the limit
$$
\lim_{x\to0^+}\frac{\sin x}{\ln\frac{1}{x}-\ln(\frac{1}{x}-1)}=
\lim_{x\to0^+}-\frac{\sin x}{\ln(1-x)}
$$
exists and they'll be equal.
