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

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?

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    $\begingroup$ Please use MathJax. $\endgroup$ – José Carlos Santos Jul 15 '17 at 9:50
  • $\begingroup$ Im on phone I cant see what it will type $\endgroup$ – Lola Jul 15 '17 at 9:53
  • $\begingroup$ I think you mean $\frac {sin 1+ sin \frac {1}{2}+\cdots+ sin \frac {1}{n}}{ln n}$ $\endgroup$ – shwetha Jul 15 '17 at 10:28
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$$\ \lim_{n\to+\infty}\frac{\sin(1)+\sin(\frac{1}{2})+\dots+\sin(\frac{1}{n})}{\ln(n)}=\lim_{n\to+\infty}\frac{\sum_{k=1}^n\sin(\frac{1}{k})}{\ln(n)}$$

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 $\ln(1+x)\sim x$

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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.

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    $\begingroup$ why is the last limit equal with$$\lim_{x\to0^+}-\frac{\sin x}{\ln(1-x)} $$ $\endgroup$ – Lola Jul 15 '17 at 11:26
  • $\begingroup$ @Lola $\ln(\frac{1}{x}-1)=\ln\frac{1-x}{x}=\ln(1-x)+\ln\frac{1}{x}$ $\endgroup$ – egreg Jul 15 '17 at 12:06

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