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Cosecant residues sum an alternating rational series over the integers

Statement

Let $f$ be a rational function such that $f (n)$ is defined for every $n \in Z$ and $f (z) = O (z^{- 2})$ as $∣ z ∣ \to \infty$ . Then

$\sum_{n \in Z} (- 1)^{n} f (n) = - \sum_{a \notin Z} Res (π \csc (π z) f (z), a),$

where the sum on the right is over the nonintegral poles of $f$ .

Facts & Assumptions

Given: A rational function $f$ with no integer pole and with $f (z) = O (z^{- 2})$ at infinity.

[L1]

The zeros of $\sin (π z)$ are exactly the integers, and $(\sin (π z))^{'} = π \cos (π z)$ (The zeros of complex sine are the integer multiples of pi, and the zeros of complex cosine are the odd half-integer multiples of pi, Complex sine, cosine, hyperbolic sine, and hyperbolic cosine are entire with their standard derivatives).

[L2]

Residues at simple zeros are computed by the quotient rule (Residues of p over q at a simple zero of q).

[L3]

The residue theorem applies on expanding rectangles (The residue theorem for a null-homologous cycle).

Proof

technique · direct

1.1

Let $F (z) = π \csc (π z) f (z)$ . By [L1], $\sin (π z)$ has a simple zero at [L1, L2, algebra] each integer $n$ , and gives $Res (F, n) = \frac{π f (n)}{π \cos (π n)} = (- 1)^{n} f (n) .$

1.2

On the same rectangles used for the cotangent theorem, the factor $\csc (π z)$ is uniformly bounded on the vertical sides because $∣ \sin (π (N + \frac{1}{2} + i y)) ∣ = ∣ \cosh (π y) ∣ \geq 1$ , and on the horizontal sides it decays exponentially like $e^{- π ∣ N ∣}$ . Since $f (z) = O (z^{- 2})$ , the boundary integral of $F$ tends to $0$ .

given

2.1

Applying [L3] and letting the rectangle expand gives $\sum_{n \in Z} (- 1)^{n} f (n) + \sum_{a \notin Z} Res (π \csc (π z) f (z), a) = 0.$ Rearranging yields the stated alternating summation formula.

step 1.1step 1.2L3∎

Depends on

Used by

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Sources

L. V. Ahlfors, Complex Analysis, 3rd ed., Ch. 4 §5.3 (standard reference, not scraped)