# Does the series $\sum_{n\ge0}\frac{x^n\sin({nx})}{n!}$ converge uniformly on $\Bbb R$?

The series $$\sum_{n\ge0}\frac{x^n\sin({nx})}{n!}$$ converges uniformly on each closed interval $[a,b]$ by Weierstrass' M-test because $$\left|\frac{x^n\sin({nx})}{n!}\right|\le\frac{\max{(|a|^n,|b|^n)}}{n!}.$$

But does this series converge uniformly on $\Bbb R$?

To rephrase Paul's answer: if the series converges uniformly, then the sequence of functions $(f_n)_{n\geq1}$ with $f_n(x)=\dfrac{x^n\sin(nx)}{n!}$ converges uniformly to zero.

If $n\geq1$, let $\xi_n=\pi\left(2n+\frac1{4n}\right)$. Then $\xi_n>n$, so that $(\xi_n)^n>n!$ and $\sin(n\xi_n)= \sin \left(\pi \left(2n^2+\tfrac14 \right)\right)=1$: we see that $f_n(\xi_n)>1$.

It follows at once that the sequence $(f_n)_{n\geq1}$ does not converge uniformly to zero.

This does not converge uniformly. For arbitrarily large $M$, there always exists an $x>M$ for which there are infinitely many values of $nx$ such that $|\sin(nx)| > 1/2$.

In particular, given $N$, there will be such values $x > N$ and thus $|\frac{x^N\sin(Nx)}{N!}|> 1/2$

So the idea is you don't have the control you need for uniform convergence.

• You mean: there exist infinitely many $n$ such that $|\sin(nx)|>1/2$, presumably. – Mariano Suárez-Álvarez May 19 '14 at 4:56