# Calculate $\lim_{n \to \infty}U_n$

How can I calculate $\lim_{n \to \infty} U_n$ where $$U_n = \sum_{k=1}^n k \sin\left(\frac{1}{k} \right)?$$

• This is at least the second question you've asked of the same type, and in both this and the other, you've shown nothing in the way of your own work, or your own thoughts, or any explanation of what you do not understand/where you're stuck. Please edit your post to include some of your own thoughts, or elaborate on where you're stuck. If you plan to frequent this site, I'd like you to know what the expectations are when asking questions, if only to spare you an onslaught of downvotes! – Namaste Dec 14 '13 at 16:17
• math.stackexchange.com/questions/390115/… – lab bhattacharjee Dec 15 '13 at 3:46
• @amWhy: aren't SOPs allowed? – Don Larynx Dec 15 '13 at 17:34
• @Don What are SOPs? Statement of the problem? Of course we want the problem in question stated, and clearly so! But I also like to encourage users to also include their thoughts, what they've tried, or if totally stuck, at least say so, and where are they stuck...etc. It's not so much of an issue with me, but I hate to see new users get bombarded with down-votes without necessarily knowing why. – Namaste Dec 15 '13 at 17:36

Applying the inequality $\sin(x) \geq \frac{2}{\pi}x$, which holds for $0 \leq x \leq \frac{\pi}{2}$, with $x = \frac{1}{k}$: $$\sum_{k=1}^n k \sin\left(\frac{1}{k}\right) \geq \sum_{k=1}^n\frac{2}{\pi} = \frac{2n}{\pi} \xrightarrow[n\to\infty]{} +\infty.$$

Remark. This estimate is actually very crude because it does not take into account the fact that $\dfrac{1}{k}$ comes closer to $0$ as $k$ grows. To go a step further, we can use the Taylor expansion of $\sin(1/k)$ and prove that $U_n-n$ converges at rate $\mathrm{O}(1/n)$ to $$C = \sum_{k=1}^\infty (-1)^k\frac{\zeta(2k)}{(2k+1)!}\approx-.2653354.$$

• @Siméon. Nice answer and nice problem. What is amazing is to see that U(n) varies almost just as n : U(10)=9.75052, U(100)=99.7363, U(1000)=999.735, U(10000)=9999.73, U(100000)=99999.7. – Claude Leibovici Dec 14 '13 at 16:29
• @ClaudeLeibovici That should be expected -- remember that $\sin(\frac{1}{k})\approx\frac{1}{k}$ for large $k$, so that you are adding up terms that are approaching $1$. – Nick Peterson Dec 14 '13 at 17:03
• @NicholasR.Peterson. For sure, but impressed me is how quickly it approaches n. Cheers. – Claude Leibovici Dec 14 '13 at 17:08
• @ClaudeLeibovici: I thought it was useful to make your observation more precise in my original answer. See my edit. – Siméon Dec 15 '13 at 16:26
• +1. But even though the sequence $n-U_n$ does converge, it is not to $\pi^2/36$. It would if one had $\sin(x)=x-x^3/6$ for every $x$ in $(0,1)$, which is not true. – Did Dec 15 '13 at 16:56

Note that for $k>0$ we have $$k\sin\left(\frac1k\right)=\frac{\sin\left(\frac1k\right)}{\frac1k},$$ and as $k\to\infty$ we have $\frac1k\to0$. Therefore $\frac{\sin\left(\frac1k\right)}{\frac1k}\to1$, by a classical result.

If so, the general term in the sum $\sum\limits_{k=1}^\infty k\sin\left(\frac1k\right)$ approaches $1$, rather than $0$, and therefore the sum, being $\lim_{n\to\infty} U_n$, must be infinite.

• Since analysis is not my field in general, I'd like to know if that answer is indeed erroneous as the downvote indicates. – Asaf Karagila Dec 17 '13 at 0:33