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Prove that the function $f$ defined by $f(x)=\sum_{n=0}^\infty\frac{\cos (n^2x)}{2^{nx}}$ is continuous on $(0,\infty)$.

I can apply the Weierstrass M-test on $[1,\infty):$ $$|\frac{\cos (n^2x)}{2^{nx}}|\le|\frac{1}{2^{nx}}|\le |\frac{1}{2^n}|$$ for $x\ge 1$ and conclude that $f$ is continuous on $[1,\infty)$ since the series is uniformly convergent there and $\frac{\cos (n^2x)}{2^{nx}}$ is continuous.

But how do I prove that $f$ is continuous on the whole $(0,\infty)$?

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    $\begingroup$ How about proving it on $[\epsilon,\infty)$? $\endgroup$ – Lord Shark the Unknown Jun 17 '18 at 11:17
  • $\begingroup$ @LordSharktheUnknown How do I prove it on $[\epsilon, \infty)$? The inequality in the Weierstrass M-test does not hold for $\epsilon < 1$ as far as I understand. $\endgroup$ – user557902851 Jun 17 '18 at 11:25
  • $\begingroup$ In this case the inequality will depend on $\varepsilon$. For $x\in[1,\infty)$ you used $x\geq1$. In this case you should use something similar $\endgroup$ – Uskebasi Jun 17 '18 at 11:42
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Take $\delta>0$. Then$$x>\delta\implies\left|\frac{\cos(n^2x)}{2^{nx}}\right|<\frac1{2^{\delta n}}$$and therefore, by the Weierstrass $M$-test your series converges uniformly on $(\delta,+\infty)$ and therefore its sum is continuous. So, if you fix $x_0\in(0,+\infty)$, then your function is continous at $x_0$, since if you take, say $\delta=\frac{x_0}2$, you know that the restriction of the function to $(\delta,+\infty)$ is continuous.

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  • $\begingroup$ That's probably too basic question, but why is $\sum \frac{1}{2^{\delta n}}$ convergent? $\endgroup$ – user557902851 Aug 9 '18 at 23:21
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    $\begingroup$ @user5579085 It's a geometric series with ratio $\frac1{2^\delta}<1$. $\endgroup$ – José Carlos Santos Aug 9 '18 at 23:24

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