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Is $\sum_{n=1}^{\infty}{(\frac{1}{n})^n}$

1)convergent aand 2) uniformly convergent?

where $x \in (-\pi,\pi)$

I was thinking to use Weierstrass criterion to show uniform convergence by getting an upper bound which converges,

$\frac{x}{n} < \frac{\pi}{n}$

$\frac{1}{n}^n < \frac{1}{n}$.

$(\frac{x}{n})^n<(\frac{\pi}{n})^n < \pi ^n . \frac{1}{n}$

But i could not find the bound?.Any other way I can proceed?

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  • $\begingroup$ problem in title doesn't match problem in question. I suspect you either mean $\sum \frac{x^n}{n}$ or $\sum\left(\frac{x}{n}\right)^n$ but can't be sure :) $\endgroup$ – N8tron Jun 12 '18 at 5:21
  • $\begingroup$ Thanks for pointing out! $\endgroup$ – BAYMAX Jun 12 '18 at 5:22
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$|\frac x n|^{n} \leq |\frac {\pi} n|^{n} \leq (\frac 1 2 )^{n}$ for $n$ sufficiently large. Compare with $\sum (\frac 1 2 )^{n}$

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  • $\begingroup$ Hm, like after $n \geq 4$, $\frac{\pi}{n}<1$ but how could we be more specific about $1/2$, after a bit of calculator type we get $\frac{\pi}{n} < \frac{1}{2} $ for $n \geq 7$, can you mention your motivation of how you could see that $\frac{1}{2}$ coming? $\endgroup$ – BAYMAX Jun 12 '18 at 5:36
  • $\begingroup$ To use M test you just have to compare with $\sum r^{n}$ for some $r \in (0,1)$ There is nothing special about $r=\frac 1 2$ but this value of $r$ works. $\endgroup$ – Kavi Rama Murthy Jun 12 '18 at 5:38
  • $\begingroup$ Nice, thanks for the idea!! $\endgroup$ – BAYMAX Jun 12 '18 at 6:26

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