Show that (sum from 1 to infinity) sin(x/n^2) converges uniformly on any closed interval [a, b].
I think I can use Weierstrass M-Test to prove this, but I can't find the limit for fn(x).
Thx in advance...
The Weiestrass M-test can be used in this case. If $\displaystyle x \in [a,b]$ and we set $\displaystyle c= max [|a|,|b|]$ is…
$\displaystyle |\sin (\frac{x}{n^{2}})|\le \frac{c}{n^{2}}$
But the series…
$\displaystyle \sum_{n=1}^{\infty} \frac {c}{n^{2}}$
… is absolutely convergent so that the series…
$\displaystyle \sum_{n=1}^{\infty} \sin (\frac{x}{n^{2}})$
… is uniformly convergent in $\displaystyle [a,b]$…
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$