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?

• Please use MathJax. – José Carlos Santos Jul 15 '17 at 9:50
• Im on phone I cant see what it will type – Lola Jul 15 '17 at 9:53
• I think you mean $\frac {sin 1+ sin \frac {1}{2}+\cdots+ sin \frac {1}{n}}{ln n}$ – shwetha Jul 15 '17 at 10:28

$$\ \lim_{n\to+\infty}\frac{\sin(1)+\sin(\frac{1}{2})+\dots+\sin(\frac{1}{n})}{\ln(n)}=\lim_{n\to+\infty}\frac{\sum_{k=1}^n\sin(\frac{1}{k})}{\ln(n)}$$
$$\ \lim_{n\to+\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\lim_{n\to+\infty}\frac{\sum_{k=1}^{n+1}\sin(\frac{1}{k})-\sum_{k=1}^n\sin(\frac{1}{k})}{\ln(n+1)-\ln(n)}= \lim_{n\to+\infty}\frac{\sin(\frac{1}{n+1})}{\ln(n+1)-\ln(n)}=$$ $$\ =\lim_{n\to+\infty}\frac{\sin(\frac{1}{n+1})}{\ln(1+\frac{1}{n})}\sim \lim_{n\to+\infty}\frac{\frac{1}{n+1}}{\frac{1}{n}}=\lim_{n\to+\infty}\frac{n}{n+1}=1$$
Where I used the fact that for $\ x\to 0$, $\sin(x)\sim x$, and $\ln(1+x)\sim x$
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.
• why is the last limit equal with$$\lim_{x\to0^+}-\frac{\sin x}{\ln(1-x)}$$ – Lola Jul 15 '17 at 11:26
• @Lola $\ln(\frac{1}{x}-1)=\ln\frac{1-x}{x}=\ln(1-x)+\ln\frac{1}{x}$ – egreg Jul 15 '17 at 12:06