Let $\displaystyle x_n=\sum_{k=1}^n\frac{k \sin^2k}{n^2+k \sin^2k}$ for all $n>0$. How can we prove that $(x_n)$ is convergent?
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What have you tried so far? What are your thoughts on the problem? (Note that boundedness of the sum is easy - each term is bounded from above by $\frac{k}{n^2}$.) – Steven Stadnicki Aug 01 '17 at 04:42
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@StevenStadnicki Yes. $x_n$ is bounded. it is easy to verify. Then monotone? or using Cauchy's criterion? I have no idea. – xldd Aug 01 '17 at 04:48
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After doing some mathematica, it seems like it converges to $.25$. – mathworker21 Aug 01 '17 at 06:04
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Clearly $\sum_{k=1}^n \frac{k\sin^2k}{n^2+n} \le x_n \le \sum_{k=1}^n \frac{k\sin^2k}{n^2}$. The result follows from observing $\sum_{k=1}^n k\sin^2k = \int_1^n x\sin^2xdx + O(n) = \frac{n^2}{4}+O(n)$.
I used the following.
Indeed, it suffices to show $\int_1^n (t-[t])(\sin^2t+t\sin(2t))dt = O(n)$, so it suffices to show $\int_1^n (t-[t])t\sin(2t)dt = O(n)$. I leave this to you.
mathworker21
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Yea, the idea is that the denominator really doesn't matter for $n$ large (since $n^2 \gg n \ge k\sin^2k$), so Squeeze theorem has to work. – mathworker21 Aug 01 '17 at 07:00
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1@mathworker21 How to prove that $\sum_{k=1}^n k \sin^2k =\int_1^n x\sin^2x dx+O(n)$? – xldd Aug 01 '17 at 10:18
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@mathworker21: This doesn't work for $\sum\limits_{k=1}^n k\sin^2(\pi k) = \int_1^n x\sin^2(\pi x),\mathrm{d}x + O(n)$ – robjohn Aug 01 '17 at 12:55
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To prove that the limit is $\frac{1}{4}$, we just need to prove that
$\sum_{k=1}^n k \sin ^2 (k) = \frac{n^2}{4} + o(n^2)$. In fact :
$\sum_{k=1}^n k \sin ^2 (k) = \sum_{k=1}^n \frac{k}{2} (1 - \cos(2k)) = \frac{n(n+1)}{4} - \frac{1}{2} \sum_{k=1}^n k\cos(2k)$.
Let now $f$ be the function defined by
$f(x) = \sum_{k=0}^n e^{ikx} = \frac{1 - e^{inx}}{1 - e^{ix}} = \frac{\sin(\frac{n}{2}x)}{\sin(\frac{x}{2})}e^{i\frac{n}{2}x}$ for $x\in\mathbb{R}\backslash 2\pi\mathbb{Z}$. It is clear that
$\sum_{k=1}^n k \sin ^2 (k) = \frac{n(n+1)}{4} - \Im(f'(1)) = \frac{n^2}{4} + o(n^2)$ since $f'(1) = O(n)$
Kroki
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$\sum_{k=1}^n e^{ikx} = \frac{1 - e^{inx}}{1 - e^{ix}}?$ I think you need $e^{ix}$ in front. – zhw. Aug 01 '17 at 17:44
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$$ \begin{align} \sum_{k=1}^n\frac{k\sin^2(k)}{n^2+k\sin^2(k)} &=n-\sum_{k=1}^n\frac{n^2}{n^2+k\sin^2(k)}\\ &=n-\sum_{k=1}^n\frac1{1+\frac{k\sin^2(k)}{n^2}}\\ &=n-\sum_{k=1}^n\left(1-\frac{k\sin^2(k)}{n^2}+O\!\left(\frac{k^2}{n^4}\right)\right)\\ &=\frac1{n^2}\sum_{k=1}^nk\sin^2(k)+O\!\left(\frac1n\right)\\ &=\frac1{n^2}\sum_{k=1}^nk\,\frac{1-\cos(2k)}2+O\!\left(\frac1n\right)\\ &=\frac14+O\!\left(\frac1n\right) \end{align} $$ Since $$\newcommand{\Re}{\operatorname{Re}} \begin{align} \sum_{k=1}^nk\cos(2k) &=\Re\left(\sum_{k=1}^nke^{2ik}\right)\\ &=\Re\left(\tfrac{e^{2i(n+1)}\left(ne^{2i}-n-1\right)+e^{2i}}{\left(1-e^{2i}\right)^2}\right)\\[6pt] &=O(n) \end{align} $$
robjohn
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I had deleted my answer because it was close to Youem's answer. However, I undeleted it since it provides some details for Youem's answer. – robjohn Aug 01 '17 at 12:51