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How can i prove uniform convergence on $E=[0, \frac{\pi}{2}]$ ? $$\sum^\infty _{n=1} {x e^{-nx}\cos(nx)}$$

Ilya Fedorov
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  • Please share your thoughts and efforts. – Servaes Oct 19 '18 at 19:08
  • I tried to use Dirichlet test for uniform convergence. $u_n(x) = xe^{-nx}$, $v_n(x) = \cos(nx)$. I dont know how to prove that $\sum^k_{n=1}{\cos(nx)}$ is bounded for any $x \in E$. $|\cos(x) + \cos(2x) + ... + \cos(nx)| <= \frac{1}{|sin(\frac{x}{2})|}$ But if $x$ is near to $0$, than $\frac{1}{|sin(\frac{x}{2})|}$ is very big, so i cant prove that this value is bounded. – Ilya Fedorov Oct 19 '18 at 19:12
  • Why do you think it converges uniformly there? I would guess it doesn't. – zhw. Oct 19 '18 at 19:36
  • Have you tried the Weierstrass M test? – Matthew C Oct 19 '18 at 19:49
  • $M-$ test doesn’t work here. You obviously can’t find an upper-bound of the absolute value due to $e^{-nx}x$. – Thinking Oct 19 '18 at 19:51
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    Define $f_N(x) = \sum_{n=1}^N xe^{-nx}\cos(nx)$ and note that $f_N(1/N) = \sum_{n=1}^N \frac{1}{N} e^{-n/N}\cos(n/N)$ is a Riemann sum. Compare $\lim_{N\to \infty}f_N(1/N)$ to $\lim_{N\to\infty}f_N(0)$. – Winther Oct 19 '18 at 20:03

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Hint: For $N$ even, let

$$T_N(x)=\sum_{n=N/2}^{N}xe^{-nx}\cos (nx).$$

If the convergence is uniform, then $T_N\to 0$ uniformly. Consider $T_N(1/N).$

zhw.
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    should be $$cos (nx).$$ instead of $$cos (Nx).$$ ? – PilouPili Oct 20 '18 at 09:04
  • @NPE Yes indeed, thank you. – zhw. Oct 20 '18 at 15:48
  • so the convergence is non-uniform, right? By the way, is the series even pointwise convergent? And, The test you display is Cauchy-Criterion right? – vidyarthi Jan 11 '19 at 17:23
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    @vidyarthi the answers are yes, yes, and on the last one I would say it shows CC does not hold.. For pointwise convergence: $x=0$ is obvious,if $x>0$ the terms in absolute value are bounded above by $xe^{-nx}.$ – zhw. Jan 11 '19 at 19:07