Alternate proof of this identity for the generalized Harmonic number?

Question

I discovered this identity starting with a continued fraction representation of the generalized harmonic number and using the alternating sum definition described in Wikipedia (Theorem 3; Corollary 3). $$H_{2k+1,r}=\cfrac{1}{1-\sum_{n=1}^k\left(\cfrac{\cfrac{1}{(2n)^r}+\cfrac{1}{(2n+1)^r}}{H_{2n-1,r}H_{2n+1,r}}\right)}$$ I was wondering if there is another way that we could prove this.

Answer 1

The sum in the denominator can be written as

$$\sum_{n=1}^k \left(\cfrac{\cfrac{1}{(2n)^r}+\cfrac{1}{(2n+1)^r}}{H_{2n-1,r}H_{2n+1,r}}\right) \\ = \sum_{n=1}^k \left( \cfrac{ H_{2n+1,r} -H_{2n-1,r} } {H_{2n-1,r}H_{2n+1,r}}\right) \\ = \sum_{n=1}^k \left( \frac {1}{H_{2n-1,r}} - \frac {1}{H_{2n+1,r} } \right) \\ = \left( \frac {1}{H_{1,r}} - \frac {1}{H_{2k+1,r} } \right) $$ $$ = 1- \frac {1}{H_{2k+1,r} } $$

Substituting this in the initial equation reported by the OP, we directly get the trivial equation

$$H_{2k+1,r}=\cfrac{1}{1- \left(1- \frac {1}{H_{2k+1,r} } \right) } = \cfrac {1}{\frac {1}{H_{2k+1,r} } } =H_{2k+1,r} $$