<s>[INST] 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?
 [/INST] 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 $f'(1) = O(n)$
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