2
$\begingroup$

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 $\Bbb R$?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

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

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.