Is $\displaystyle\sum_{k=1}^n \frac{k \sin^2k}{n^2+k \sin^2k}$ convergent? 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?
 A: 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.
A: 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 $f'(1) = O(n)$
A: $$
\begin{align}
\sum_{k=1}^n\frac{k\sin^2(k)}{n^2+k\sin^2(k)}
&=n-\sum_{k=1}^n\frac{n^2}{n^2+k\sin^2(k)}\\
&=n-\sum_{k=1}^n\frac1{1+\frac{k\sin^2(k)}{n^2}}\\
&=n-\sum_{k=1}^n\left(1-\frac{k\sin^2(k)}{n^2}+O\!\left(\frac{k^2}{n^4}\right)\right)\\
&=\frac1{n^2}\sum_{k=1}^nk\sin^2(k)+O\!\left(\frac1n\right)\\
&=\frac1{n^2}\sum_{k=1}^nk\,\frac{1-\cos(2k)}2+O\!\left(\frac1n\right)\\
&=\frac14+O\!\left(\frac1n\right)
\end{align}
$$
Since
$$\newcommand{\Re}{\operatorname{Re}}
\begin{align}
\sum_{k=1}^nk\cos(2k)
&=\Re\left(\sum_{k=1}^nke^{2ik}\right)\\
&=\Re\left(\tfrac{e^{2i(n+1)}\left(ne^{2i}-n-1\right)+e^{2i}}{\left(1-e^{2i}\right)^2}\right)\\[6pt]
&=O(n)
\end{align}
$$
