Help me please to compute this limit:

$\lim\limits_{n \to \infty} \frac{1}{n}\left ( \cos1+\cos(\frac{1}{2})+\cos(\frac{1}{3}) +...+\cos (\frac{1}{n}) \right )$

Thank you.

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    $\begingroup$ Do you know that for each convergent sequence $x_n \to x_\infty$ also the sequence $(x_1+\cdots+x_n)/n$ of arithmetic means converges to $x_\infty$? $\endgroup$ – Jochen Jan 28 '13 at 9:46
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    $\begingroup$ Hint. Recall that, as for any alternating series, we can write $1-\frac{x^2}{2}\leq \cos(x)\leq 1-\frac{x^2}{2}+\frac{x^4}{4!}$ $\endgroup$ – uforoboa Jan 28 '13 at 9:46
  • $\begingroup$ @Sanchez: I disagree. A Riemann sum that would seem to be useful here is of the form $\frac{1}{n} \sum_{k=0}^{n} f(k/n)$; we do not have that here. Nor do we have anything for a varying interval, as the sum is not weighted. $\endgroup$ – Ron Gordon Jan 28 '13 at 9:51
  • $\begingroup$ @rlgordonma, wow you are quick. I thought I deleted that comment within 15s I posted it. $\endgroup$ – user27126 Jan 28 '13 at 10:22
  • $\begingroup$ I guess you did. Still useful to demonstrate that it is not a Riemann sum despite appearances. Glad you realized it. $\endgroup$ – Ron Gordon Jan 28 '13 at 11:48

Define $$a_n:=\sum_{j=1}^{n}\cos\left(\frac{1}{j}\right)$$ and $$b_n:=n.$$ Then notice that $$\lim_{n\to\infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\lim_{n\to\infty}\cos \left(\frac {1}{n+1}\right)=1,$$ and therefore by Stolz Cesaro Lemma also your limiti exists and it is equal to $1$.


We may nicely squeeze it $$\lim\limits_{n \to \infty}\sqrt[n]{\cos (1)\times\cos (1/2)\times\cdots\times\cos(1/n)}\le\lim\limits_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n}\cos\left(\frac{1}{k}\right)\le\lim\limits_{n \to \infty}\left(\frac{1}{n}\times n\cos(1/n)\right)$$ $$1\le\lim\limits_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n}\cos\left(\frac{1}{k}\right)\le1$$ where on the left side I combined AM-GM and Cauchy-d'Alembert criterion.


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