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I was doing a problem where I was asked to show the continuity of $$\sum_{n = 1}^{\infty} e^{-nx}\sin{nx},$$ for $x>0$.

My approach was to consider the sequence $(\sigma_{n}(x))$ of partial sums of the series $$\sum_{n = 1}^{\infty} |e^{-nx}\sin{nx}|,$$ to get $$\sigma_{n}(x) = \frac{e^{-x}(1-e^{-nx})}{1-e^{-x}}.$$ Which converges to the function $$ \sigma(x) := \frac{e^{-x}}{1-e^{-x}},$$ uniformly on $\{x > 0\}$. This implies that the given series converges absolutely to $\sigma(x)$ on the given set.

How do I say from here that the given series is also uniformly convergent?

Also, is it necessary that if a series of functions converges absolutely, then it also converges uniformly? Am I missing this in the above proof?

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    $\begingroup$ $\sum \frac x {n^{2}}$ is absolutely convergent on $(0,\infty)$ but not uniformly convergent. $\endgroup$ Commented Sep 8 at 8:31
  • $\begingroup$ @KaviRamaMurthy Nice counterexample. Thanks! $\endgroup$ Commented Sep 8 at 8:41

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Apply M-test to show uniform convergence on $[a,\infty)$ for any $a>0$: Just use the fact that $|\sin (nx)|\le 1$ and $\sum e^{-na}<\infty$. Hence, the sum is continuous on $[a,\infty)$ for any $a>0$ which proves continuity on $(0,\infty)$.

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  • $\begingroup$ This type of estimate never came to my mind. Thanks! $\endgroup$ Commented Sep 8 at 8:42
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For $$f(x)=\sum_{n = 1}^{\infty} e^{-nx}\sin(nx)$$ why not to use Euler represntation of the sine function which leads to $$f(x)=\frac {\sin(x)}{2(\cosh(x)-\cos(x))}$$ which is continuous everywhere except at $x=0$.

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  • $\begingroup$ Thanks! I will research more on this. $\endgroup$ Commented Sep 8 at 12:57
  • $\begingroup$ @user1684451. This is just a geometric series. Start with $$=\sum_{n = 1}^{\infty} e^{-nx}\,e^{inx)}=e^{(1+i) n x}=\frac{e^{(1+i) x}}{1-e^{(1+i) x}}$$ Its imaginary part is $$\frac{e^x \sin (x)}{e^{2 x} \sin ^2(x)+\left(1-e^x \cos (x)\right)^2}=\frac {\sin(x)}{2(\cosh(x)-\cos(x))}$$ $\endgroup$ Commented Sep 10 at 11:53
  • $\begingroup$ ohh, this makes sense now! $\endgroup$ Commented Sep 13 at 7:23

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