# Finding the sum of a certain series

Consider $\sum_{n=1}^{\infty}nx^n\sin(nx)$. Find $R > 0$ such that the series is convergent for all $x\in(-R,R)$. Calculate the sum of the series.

I could find the radius of convergence is $R=1$, hence for any $x\in (-1,1)$ the series is continuous and convergent, However, I have some problem in finding the exact sum of this series.

To find $f(x)=\sum_{n=1}^{\infty}nx^n\sin(nx)$, I think it's reasonable to find $F(x)=\sum_{n=1}^{\infty}nx^ne^{inx}$ and the imaginary part of $F(x)$ is $f(x)$.

So if $F(x)=\sum_{n=1}^{\infty}nx^ne^{inx}$, then $\frac{1}{2\pi}\int_{-\pi}^\pi F(x)e^{-inx}dx=nx^n$, but I don't know how to find $F(x).$

• You can generalize your sum and try to compute $\sum_{n=1}^\infty n \alpha^n$. Do you know how to do that? – Idéophage Aug 2 '17 at 3:16
• @Idéophage not really! – Parisina Aug 2 '17 at 4:00
• Do you know how to compute $\sum_{n=1}^\infty \alpha^n$? – Idéophage Aug 2 '17 at 4:51

Your idea to replace $\sin(nx)$ by $e^{inx}$ is good. Now you can write $\sum_{n=0}^\infty n x^n e^{inx} = \sum_{n=0}^\infty n (x e^{ix})^n$ and let $\alpha := x e^{ix}$. Our goal is to compute $$\sum_{n=0}^\infty n \alpha^n \text{.}$$

### First way

Remark that the sum is \begin{align*} \alpha^1 + \alpha^2 + \alpha^3 + \cdots\\ \phantom{\alpha^1} + \alpha^2 + \alpha^3 + \cdots\\ \phantom{\alpha^1 + \alpha^2} + \alpha^3 + \cdots\\ \phantom{\alpha^1 + \alpha^2 + \alpha^3} + \cdots\\ \end{align*}

You probably know a formula for each line.

Alternatively, you can see that the sum is $(1 + \alpha + \alpha^2 + \cdots)(\alpha^1 + \alpha^2 + \alpha^3 + \cdots)$. Yet an other way to present that is to multiply the sum by $1-\alpha$ (“discrete differentiation”): we get $\sum_{n=1}^\infty \alpha^n$ (multiplying again by $1-\alpha$ gives $\alpha$). If you continue on that idea, you see Pascal's triangle appearing and it gives you a way to compute sums of the form $\sum_{n=0}^\infty P(n) \alpha^n$ where $P$ is a polynomial.

### Second way

We know that $\sum_{n=0}^\infty \alpha^n = \frac{1}{1-\alpha}$. Now take the derivative of each side of the equality.

The derivative of the well-known geometric series is $$\sum^\infty_{n=1}n\alpha^{n-1}=\frac{d}{d\alpha}\frac1{1-\alpha}=\frac1{(1-\alpha)^2},$$ this means $$\sum^\infty_{n=1}n\alpha^n=\frac{\alpha}{(1-\alpha)^2}.$$ Now using $\sin nx=\frac1{2i}(e^{inx}-e^{-inx}),$ we obtain (with $\alpha=xe^{ix}$ and $\alpha=xe^{-ix}$) $$\sum^\infty_{n=1}nx^n\sin nx=\frac1{2i}\left(\frac{xe^{ix}}{(1-xe^{ix})^2}-\frac{xe^{-ix}}{(1-xe^{-ix})^2}\right).$$ The RHS can be simplified: $$\frac1{2i}\frac{xe^{ix}(1-xe^{-ix})^2-xe^{-ix}(1-xe^{ix})^2}{(1-xe^{ix})^2(1-xe^{-ix})^2}=\frac{x(1-x^2)\sin x}{(1-2x\cos x+x^2)^2}.$$ An alternative (staying in the area of real functions) is to use the generating function of Chebyshev polynomials of first kind (http://mathworld.wolfram.com/ChebyshevPolynomialoftheFirstKind.html): $$\frac{1-t^2}{1-2xt+t^2}=1+2\sum^\infty_{n=1}T_n(x)t^n$$ for $|t|<1$. Replacing $x$ by $\cos x,$ this becomes $$\frac{1-t^2}{1-2t\cos x+t^2}=1+2\sum^\infty_{n=1}t^n\cos nx .$$ Differentiating with respect to $x$ (this is justified, because the resulting series is uniformly convergent for $|t|<1$) and setting $t=x$, we arrive at the same result.