# Why do Candelas and Howard say that $\sum_{n=1}^\infty \cos\left( n \kappa \epsilon \right) \ = \ - \frac{1}{2}$?

In the paper Vacuum $$\langle \phi^2 \rangle$$ in Schwarzschild Spacetime by Candalas and Howard, they say that for each non-zero $$\epsilon$$ it is true that $$\sum_{n=1}^\infty \cos\left( n \kappa \epsilon \right) \ = \ - \frac{1}{2}$$

This is equation (2.7) in the paper, where $$\kappa$$ is a constant (later set as the surface gravity for the black hole) and $$\epsilon \to 0^{+}$$ is taken as a regulator.

In what sense is this true? As some kind of distributional statement? Because this doesn't converge in the strict sense.

• Compute $\sum_{n=1}^\infty e^{-n\rho} \cos (nx)$ instead and then take $\rho \to 0^+$ which gives $-\frac{1}{2}$. At this stage you can take $\epsilon \to 0^+$. – Prahar May 15 '19 at 2:28

Note that $$III(x)~=~ \delta(x-\mathbb{Z})~=~ \sum_{m\in\mathbb{Z}} \delta(x-m)~=~\sum_{n\in\mathbb{Z}}e^{2\pi i xn}~=~1+2\sum_{n\in\mathbb{N}}\cos(2\pi xn)$$ is the Dirac comb/Shah distribution.
• So for $x \notin \mathbb{Z}$ (which is going to be true for $x = \kappa \epsilon/(2\pi)$ if $\epsilon$ is small enough), then we have $\delta(x - \mathbb{Z}) =0$ and from which the identity follows? – QuantumEyedea May 15 '19 at 18:55
• $\uparrow$ Yes. – Qmechanic May 15 '19 at 18:57