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For an integer $\,r,\,$ let $$f(x)=\sum\limits_{n=1}^\infty (-1)^{rn}\,\frac{\sin nx}{n^2}.$$ Then we need to prove $f(x)$ is convergent for every $x\in\Bbb R$ and every integer $r$.

My approach is to consider $\,a(n)=\dfrac1{n^2}\,$ and $\,b(n)=(-1)^n\sin(nx).$

The series $\;\sum\limits_{n=1}^{\infty}\dfrac1{n^2}\;$ is bounded and convergent and the sequence $\;b(n)=(-1)^n\sin(nx)\,$ oscillates between positive and negative values as $\,n\,$ increases, but its magnitude is bounded.

Is the approach correct and how to incorporate $\,r\,$ in the solution?

Anne Bauval
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Vinod sharma
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1 Answers1

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Actually, $\left((-1)^n\sin(nx)\right)_{n\in\mathbb{N}}$ is a bounded sequence, for eqch $x\in\mathbb{R}$, since, for each $n\in\mathbb{N}$, $\left|(-1)^n\sin(nx)\right|\le1$. More generally, $\left((-1)^{rn}\sin(nx)\right)_{n\in\mathbb{N}}$ is a bounded sequence (the absolute value of every element of the sequence is smaller than or equal to $1$). So, $$ \left|(-1)^{nr}\frac{\sin(nx)}{n^2}\right|\le\frac1{n^2} $$ and therefore your series converges absolutely. So, it is convergent.

Another User
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