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I'm not sure wether or not the following sum uniformly converge on $\mathbb{R}$ :

$$\sum_{n=1}^{\infty} \frac{\sin(n x) \sin(n^2 x)}{n+x^2}$$

Can someone help me with it? (I can't use Dirichlet' because of the areas where $x$ is close to $0$)

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It's a tough series to analyze, in part because it doesn't seem to absolutely converge anywhere except when $x$ is a multiple of $\pi$. –  Greg Martin May 20 '13 at 23:39
    
I know, It was on a contest at my university, no one solved it (with a proof) and the professors are yet to respond my request for an answer/proof. Generally, It seems like Sum(sin(nx)/n) does not converge uniformly and this series seems "close enough" to it to not converge uniformly too, yet it obviously nothing but a very very not solid ground for speculations only. –  ORBOT Inc. May 21 '13 at 5:15
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$$\theta_3(z,q)=\sum_{n=-\infty}^\infty q^{n^2}e^{2n iz}$$ $$\frac{1}{2i}\int_{0}^{x/2}\theta_3(z,e^{2\pi i x})-\theta_3(z,e^{2\pi i x}) dz=\sum_{n=1}^\infty\frac{\sin(nx)\sin(n^2x)}{n}$$ –  Ethan May 21 '13 at 8:46
    

1 Answer 1

up vote 43 down vote accepted

The series does converge uniformly. For the proof, put $S_n(x) = \sum_{k = 0}^n \sin{(kx)}\sin{(k^2 x)}$ for $n\geq 0$. The general idea is to use summation by parts to reduce ourselves to showing that $S_n(x)$ is bounded uniformly, and then to prove that by giving a closed form for $S_n(x)$.

First, the summation by parts (I write $S_n$ in place of $S_n(x)$ for brevity): $$ \begin{align} \sum_{n = 1}^N{\sin{(nx)}\sin{(n^2 x)}\over n + x^2} & = \sum_{n = 1}^N {1\over n+x^2}(S_n - S_{n-1}) \\ & = \sum_{n = 1}^N {S_n\over n+x^2} - \sum_{n = 1}^N {S_{n-1}\over n+x^2} \\ & = \sum_{n = 1}^N {S_n\over n+x^2} - \sum_{n = 0}^{N-1} {S_{n}\over n+1+x^2} \\ & = {S_N\over N+x^2} - {S_0\over 1 + x^2} + \sum_{n = 1}^{N-1} S_n\left({1\over n+x^2} - {1\over n+1+x^2}\right) \\ & = {S_N\over N+x^2} + \sum_{n = 1}^{N-1} {S_n\over (n+x^2)(n+1+x^2)}. \end{align} $$ From here it is clear that it is enough to prove that $S_n = S_n(x)$ is uniformly bounded in $x$.

To do this, note that $$ \begin{align} 2\sin{(kx)}\sin{(k^2 x)} &= \cos{\{(k^2 - k)x\}} - \cos{\{(k^2 + k)x\}} \\ & = \cos{\{k(k - 1)x\}} - \cos{\{(k+1)k)x\}}, \end{align} $$ and therefore that $$ \begin{align} 2S_n(x) & = \sum_{k = 0}^n 2\sin{(kx)}\sin{(k^2 x)} \\ & = \sum_{k = 0}^n\left(\cos{\{k(k - 1)x\}} - \cos{\{(k+1)k)x\}}\right) \end{align} $$ The last sum telescopes, and we are left with $$ 2S_n(x) = 1 - \cos{\{n(n+1)x\}}, $$ which is plainly bounded uniformly in $x$. So we're done.

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This needs many many more upvotes! But, isn't the formula $$\sin a\sin b=\color{red}{\frac 1 2}\text{blah ? }$$ –  Pedro Tamaroff May 24 '13 at 0:41
    
@PeterTamaroff Yes it is! I just fixed it, thanks. –  Nick Strehlke May 24 '13 at 1:43

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