1
$\begingroup$

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 $F(x).$

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

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.