# Symmetry formula for Generalized Linear Euler sums

$$\sum_{k=1}^\infty \frac{H^{(p)}_k}{k^q}+\sum_{k=1}^\infty \frac{H^{(q)}_k}{k^p} =\zeta(p)\zeta(q)+\zeta(p+q)$$

$$\textit{proof}$$
Take the leftmost series and swap the finite and infinite sums

$$\sum_{i=1}^\infty \,\sum_{k=i}^\infty\frac{1}{i^p} \frac{1}{k^q}=\sum_{i=1}^\infty \,\sum_{k=1}^\infty\frac{1}{i^p} \frac{1}{k^q}-\sum_{i=1}^\infty\frac{1}{i^p} \,\sum_{k=1}^{i-1} \frac{1}{k^q}$$

The second sum can be written as

$$\sum_{i=1}^\infty\frac{1}{i^p} \,\sum_{k=1}^{i-1} \frac{1}{k^q} = \sum_{i=1}^\infty\frac{1}{i^p} \,\left(\sum_{k=1}^{i} \frac{1}{k^q}-\frac{1}{i^p}\right)$$

By separating and changing the index we get

$$\sum_{k=1}^\infty\frac{H^{(q)}_k}{k^p}-\zeta(p+q)$$

Hence we have

$$\sum_{k=1}^\infty \frac{H^{(p)}_k}{k^q} =\zeta(p)\zeta(q)-\sum_{k=1}^\infty\frac{H^{(q)}_k}{k^p}+\zeta(p+q)$$

Which reduces to
$$\sum_{k=1}^\infty \frac{H^{(p)}_k}{k^q}+\sum_{k=1}^\infty\frac{H^{(q)}_k}{k^p} =\zeta(p)\zeta(q)+\zeta(p+q)$$

For the special case $p=q=n$

$$\sum_{k=1}^\infty \frac{H^{(n)}_k}{k^n} =\frac{\zeta^2(n)+\zeta(2n)}{2}$$

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