-1

Consider the sum $$ S_n= \sum_0^n \frac{\sin^2(k)}{2k+1}.$$Is it convergent ? Can it be evaluated in the form of a closed expression for any $n \in \mathbb N$ or for large $n.$ Can we somehow use Fourier series? Thank you for any help in advance

AgnostMystic
  • 1,808
  • 2
    Every time $k$ goes around the unit circle (i.e. goes through $2\pi$ radians), we get at least one $k$ in $[2 \pi m + \pi/2 - 1/2, 2 \pi m + \pi/2 + 1/2)$ for $m$ an integer. What's a lower bound of the sum of just the terms for these $k$? – Eric Towers Apr 13 '24 at 06:53
  • Hint: to prove divergence, check $\sum_0^n\frac{\cos2k}{2k+1}$ converges. – J.G. Apr 14 '24 at 07:03
  • Doesn't Dirichlet's test suggests that the series converges? I'm not sure how to show $\sum (\sin k)^2$ is bounded beyond, saying $(\sin k)^2 \leq |\sin k|$ and $\sum (\sin k)$ is bounded. I'm not familiar with Claude's method, but if I assume he's correct, then the absolute values in my inequality must make $\sum |\sin k |$ unbounded. Clearly, the other conditions for Dirichlet's test apply. – nickalh Apr 14 '24 at 09:15
  • I found this, so I guess so. https://math.stackexchange.com/questions/127381/is-the-series-sum-sinnn-divergent?rq=1 – nickalh Apr 14 '24 at 09:17

1 Answers1

1

Too long for comments.

In terms of special functions, there is a closed form for the partial sum $S_n$ (have a look here).

But you must also notice is that $$I_n=\int_0^n\frac {\sin^2(k)} {2k+1}\,dk$$ is quite easy to compute $$I_n=\frac{1}{4} (\cos (1) (\text{Ci}(1)-\text{Ci}(2 n+1))+\sin (1) (\text{Si}(1)-\text{Si}(2 n+1))+\log (2 n+1))$$ does not converge and tends to infinity.

Now, since the integral exists, use the simplest form of Euler-MacLaurin summation formula and show that $$S_n\sim C+\frac 14 \log(n)+O\left(\frac{1}{n}\right)$$ where $C$ is a small number.

Claude Leibovici
  • 289,558