Special Series — Question 2

*Question 2 Consider $\displaystyle\sum_{n=1}^{\infty}\frac1{n^4}$.

1. Prove convergence and compare its speed with the Basel series.

2. State Euler’s classical value $\zeta(4)$.

3. Using that value, compute the even-indexed and odd-indexed fourth-power subseries separately.

*Question 2 – Solution

The series is a $p$-series with $p=4>1$, so it converges. Moreover, $1/n^4\le1/n^2$ for $n\ge1$, so its tail decays substantially faster than the Basel-series tail.

Euler’s classical evaluation is $$\zeta(4)=\sum_{n=1}^{\infty}\frac1{n^4}=\frac{\pi^4}{90} \approx1.082323.$$ As with $\zeta(2)$, the $p$-series test proves convergence but not this exact constant.

For even indices $n=2k$, $$\sum_{k=1}^{\infty}\frac1{(2k)^4} =\frac1{16}\sum_{k=1}^{\infty}\frac1{k^4} =\frac{\pi^4}{1440}.$$

Subtracting the even terms from the full series gives $$\sum_{k=1}^{\infty}\frac1{(2k-1)^4} =\frac{\pi^4}{90}-\frac{\pi^4}{1440} =\frac{\pi^4}{96}.$$ The even and odd pieces recombine to $\pi^4/90$, providing an algebraic check.