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It is known that ($N$ positive integer)

$$ \frac{1}{2N} \sum_{n=0}^{N-1} \frac{1}{\left[\cos\left( \frac{(2n+1)\pi}{4N} \right)\right]^2} = N $$

see for example this MSE post A curious identity on sums of secants.

Is there a closed form for the analogous sum where first powers are involved?

$$ S_1(N):= \frac{1}{2N} \sum_{n=0}^{N-1} \frac{1}{\cos\left( \frac{(2n+1)\pi}{4N} \right)} $$

I suspect the answer won't be entirely trivial as $S_1(N)$ diverges logarithmically as $N\to \infty$.

I would be also interested in the analogous sum where the angles are even multiples of $\pi/(4 N)$, i.e., what about

$$ Z_1(N):= \frac{1}{2N} \sum_{n=0}^{N-1} \frac{1}{\cos\left( \frac{n\pi}{2N} \right)} $$

?

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lcv
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  • @Carl I know the asymptotics (which I can compute with Euler-Maclaurin). An alternative approach (posted by D'Aurzio and now deleted) is to use the series expansion of $1/\sin(k)$ which has radius of convergence $\pi$. But my question is really if it's possible to obtain a closed form expression. – lcv Jan 12 '17 at 17:53
  • That sum appears as (part of) the susceptibility of the Ising model at the critical point. Some quantities of this model admit a closed form expression (for example the fidelity susceptibility) – lcv Jan 12 '17 at 17:56

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