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How would you go about proving that $\sum\limits_{k}\frac{1}{k}\sin\pi\left(\frac{k^2}{x+k}\right)$ converges uniformly in ($0$, $\pi/2$)

I'll appreciate any help.

I thought the Dirichlet test for uniform convergence could be used here since I can easily show that $|\frac{1}{k}\sin\pi\left(a_k(x)\right)|\leq \frac{1}{k}$ converges uniformly to 0, but I'd still need to get a bounded sequence of partial sums there somewhere (maybe manipulate the sine somehow) plus monotonicity and I'm not sure if this is even a right way to go.

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The series converges uniformly in $[0,X]$ for every finite $X$. First, we have $k^2/(k+x)=k-x+x^2/(k+x)$. Considering $\sin(k\pi-y)=(-1)^{k-1}\sin\,y$, the series becomes $$ \sum\limits_{k=1}^\infty\frac{1}{k}\,\sin\pi\,\frac{k^2}{x+k}= \sum\limits_{k=1}^\infty\frac{(-1)^{k-1}}{k}\,\sin\pi\left(x-\frac{x^2}{x+k}\right). $$ Using the addition theorem, this equals $$ \sin\pi x\,\sum\limits_{k=1}^\infty\frac{(-1)^{k-1}}{k} +\sin\pi x\,\sum\limits_{k=1}^\infty\frac{(-1)^{k-1}}{k}\left(\cos\pi\,\frac{x^2}{x+k}-1\right) -\cos\pi x\,\sum\limits_{k=1}^\infty\frac{(-1)^{k-1}}{k}\,\sin\pi\,\frac{x^2}{x+k}. $$ Because of the boundesness of $\sin$, the first series converges uniformly. So does the second series, since $|\cos y -1|=2\,\sin^2y/2\le y^2/2$, and the third one, given $|\sin y|\le |y|$.

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