Does this series converge to an actual number for any value of $x$? $$\sum_ {k = 1}^{\infty} \frac{\ln(k)\sin( 2\pi kx) }{ k}. $$ I tried summing the series for $x=2/3$ on wolfram alpha, and it seems bounded, but it also just seems to oscillate alot. Does it actually approach something? If so, does anyone know how to speed up the convergence of a sequence like this?


It does converge for all real $x$ by Dirichlet's test. However, for almost all $k$ it only converges conditionally. To speed up convergence a bit, you might try summation by parts.

EDIT: Let $$\eqalign{a_n &= \frac{\ln(n)}{n}\cr d_n &= a_n - a_{n+1} = \frac{\ln(n)}{n} - \frac{\ln(n+1)}{n+1}\cr b_n(x) &= \sin(2 \pi n x)\cr B_n(x) &= \sum_{k=1}^n b_k(x) = \frac{\sin(2 \pi n x) - \sin(2 \pi (n+1)x) + \sin(2 \pi x)}{1 - \cos(2 \pi x)}}$$

Summation by parts says

$$ \sum_{k=1}^n a_k b_k = \sum_{k=1}^{n-1} d_k B_k + a_n B_n $$

and thus if $a_n \to 0$ and $B_n$ stays bounded as $n \to \infty$,

$$ \sum_{k=1}^\infty a_k b_k = \sum_{k=1}^\infty d_k B_k $$

Here is an animation of the partial sums $\sum_{k=1}^N a_k b_k$ (red) and $\sum_{k=1}^{N} d_k B_k$ (blue) : you can see that the blue curve settles down to a limit faster than the red curve does.

enter image description here

  • $\begingroup$ Converges conditionally? What does that mean? And by summation by parts, do you mean breaking it up into congruence classes if x is rational say x=a/b, and then summing it when k is of the form bn-1,bn-2,...bn-(b-1), for any integer n? $\endgroup$ – Ethan Nov 29 '12 at 7:44
  • $\begingroup$ en.wikipedia.org/wiki/Conditional_convergence And no, I mean en.wikipedia.org/wiki/Summation_by_parts $\endgroup$ – Robert Israel Nov 29 '12 at 7:49
  • $\begingroup$ I don't understand, isnt that just a re-arangement of the summands? $\endgroup$ – Ethan Nov 29 '12 at 7:57
  • $\begingroup$ It is a kind of rearrangement, but now the terms are $O(\log(n)/n^2)$ rather than just $O(\log(n)/n)$. In particular, the series becomes absolutely convergent. $\endgroup$ – Robert Israel Nov 29 '12 at 8:14
  • $\begingroup$ Can you do the transformation for me in your anwser, I would appreciate it alot, im not sure which function to re-write so that it works most efficiently. $\endgroup$ – Ethan Nov 29 '12 at 8:17

Note that $$\sum_{k=1}^N \sin(2 \pi k x) = \csc(\pi x) \sin(N \pi x) \sin((N+1)\pi x)$$ Hence, $$\left \vert \sum_{k=1}^N \sin(2 \pi k x) \right \vert = \left \vert \csc(\pi x) \sin(N \pi x) \sin((N+1)\pi x) \right \vert \leq \left \vert \csc(\pi x) \right \vert$$ Hence, $\displaystyle \left \vert \sum_{k=1}^N \sin(2 \pi k x) \right \vert$ is bounded for all $N$.

Also, we have that if $a_k = \dfrac{\log(k)}{k}$, then $a_k$ is a monotone decreasing sequence to $0$. Now use Dirichlet's test to conclude that the series converges for all $x$.


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