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I was doing an exercise concerning uniform convergence of series of functions and I got stuck trying to finish the following:

Give an example of sequences $(a_n)$, $(b_n)$ such that

  • $\sum\limits_{k=1}^{\infty}a_k$ converges absolutely.
  • $\sum\limits_{k=1}^{\infty}f_k '$ does not converge pointwise.

Where $f_k(x):=a_k\cos(xb_k)$

I started by setting $a_n=-1/n^2$, but I cannot find $(b_n)$ such that the series of the derivatives $f_k'(x)=\frac{1}{k^2} b_k\sin(xb_k)$ diverges. My problem is that I cannot bound the $\sin$ function from below and I do not know the behavior of $\sin(xb_k)$ as $k$ changes. Indeed, fixing any $x$, its sign changes as a "messy" function of $k$.

Is my plan correct? Should I try with a different $(a_n)$?

Any hint is highly appreciated.

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    $\begingroup$ Try with $b_k=k^3$ $\endgroup$ – hamam_Abdallah Apr 22 '17 at 22:47
  • $\begingroup$ @Salahamam_Fatima I tried, but I have problems figuring out how to deal with $\sin(k^3x)$ for fixed $x$, since its sign changes as $k$ does. Intuitively, it seems to me that half of the time the $\sin$ is positive and it is negative for the rest of time, which makes divergence difficult to prove (if the sum actually diverges) $\endgroup$ – A-B-izi Apr 22 '17 at 22:52
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    $\begingroup$ Try $b_k=k^3$ and $x=\pi/2$... $\endgroup$ – sranthrop Apr 22 '17 at 23:03
  • $\begingroup$ @sranthrop Thank you very much, I think I solved it. Again I had to deal with $\sin(k^3\pi/2)$ changing sign, but at least the pattern is very simple in this case. $\endgroup$ – A-B-izi Apr 22 '17 at 23:24
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    $\begingroup$ You're very welcome :) $\endgroup$ – sranthrop Apr 22 '17 at 23:26

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