Express \[ f(z)=\sum_{n=-\infty}^{\infty}\frac1{(z-n)^4} \] in terms of trigonometric functions, and use the result to evaluate \[ \sum_{n=1}^{\infty}\frac1{n^4}. \]
Solution
1. Recall the partial-fraction identity\[ S_2(z):=\sum_{n\in\mathbb Z}\frac1{(z-n)^2} =\pi^2\csc^2(\pi z). \]
This is the Mittag–Leffler expansion of \(\pi^2\csc^2(\pi z)\). Both sides are meromorphic with double poles at the integers and the same principal parts.
2. On compact subsets of \(\mathbb C\setminus\mathbb Z\), the series for\(S_2\) may be differentiated twice term by term.
If a compact set \(K\) stays a positive distance from \(\mathbb Z\), then for large \(|n|\) and \(z\in K\) one has \(|z-n|\ge |n|/2\). The differentiated terms are bounded by constant multiples of \(|n|^{-3}\) and \(|n|^{-4}\), respectively. These majorants are summable, so the differentiated series converge uniformly on \(K\).
3. Therefore\[ 6f(z)=\frac{d^2}{dz^2}\bigl(\pi^2\csc^2(\pi z)\bigr). \]
For each summand, \[ \frac{d^2}{dz^2}(z-n)^{-2}=6(z-n)^{-4}. \] Now apply 2 to the identity in 1.
4. One obtains the trigonometric formula\[ \boxed{ f(z) =\pi^4\csc^4(\pi z) -\frac{2\pi^4}{3}\csc^2(\pi z).} \]
Put \(u=\pi z\). Since \[ \frac{d}{du}\csc^2u=-2\csc^2u\cot u, \] one more derivative gives \[ \frac{d^2}{du^2}\csc^2u =4\csc^2u\cot^2u+2\csc^4u. \] Thus 3 gives \[ f(z) =\frac{\pi^4}{6} \bigl(4\csc^2u\cot^2u+2\csc^4u\bigr). \] Using \(\cot^2u=\csc^2u-1\) reduces this to the displayed expression.
5. The Laurent expansion at \(z=0\) has constant term\[ 2\sum_{n=1}^{\infty}\frac1{n^4} \]on the series side.
Separate the \(n=0\) term: \[ f(z)=\frac1{z^4} +\sum_{n\ne0}\frac1{(z-n)^4}. \] The second sum is holomorphic near \(0\), and its value at \(0\) is \[ \sum_{n\ne0}\frac1{n^4} =2\sum_{n=1}^{\infty}\frac1{n^4}. \]
6. The constant Laurent term of the trigonometric expression in 4is \(\pi^4/45\).
The standard expansion at the origin is \[ \csc^2(\pi z) =\frac1{\pi^2z^2} +\frac13 +\frac{\pi^2z^2}{15} +O(z^4). \] Squaring gives \[ \csc^4(\pi z) =\frac1{\pi^4z^4} +\frac{2}{3\pi^2z^2} +\frac{11}{45} +O(z^2). \] Substitution into 4 yields \[ f(z) =\frac1{z^4} +\left(\frac{11}{45}-\frac{2}{9}\right)\pi^4 +O(z^2) =\frac1{z^4}+\frac{\pi^4}{45}+O(z^2). \]
7. Hence\[ \boxed{\displaystyle \sum_{n=1}^{\infty}\frac1{n^4}=\frac{\pi^4}{90}.} \]
Equating the constant terms from 5 and 6 gives \[ 2\sum_{n=1}^{\infty}\frac1{n^4}=\frac{\pi^4}{45}. \] Dividing by \(2\) proves the formula.