While playing with the Basel Problem, I derived this generalized Riemann Zeta $\zeta(s)$ formula
Proposition
$$\frac{1}{s-1}+\sum_{n=1}^{\infty}\left[\frac{1}{n^s}+\frac{1}{s-1}\left(\frac{1}{(n+1)^{s-1}} – \frac{1}{n^{s-1}}\right)\right] = \zeta(s)$$
Proof
Area below the curve is
$$\int_{1}^{\infty} \frac{1}{x^s} \, dx = \frac{1}{s-1}$$
Area above the curve for each rectangle is
$$\frac{1}{n^s} – \int_{n}^{n+1} \frac{1}{x^s} \, dx = \frac{1}{n^s}+\frac{1}{s-1}\left(\frac{1}{(n+1)^{s-1}} – \frac{1}{n^{s-1}}\right)$$
Total area above and below the curve is
$$\frac{1}{s-1}+\sum_{n=1}^{\infty}\left[\frac{1}{n^s}+\frac{1}{s-1}\left(\frac{1}{(n+1)^{s-1}} – \frac{1}{n^{s-1}}\right)\right]$$
Rearranging the series
\begin{align*}
\small \sum_{n=1}^{\infty}\left[\frac{1}{n^s}+\frac{1}{s-1}\left(\frac{1}{(n+1)^{s-1}} – \frac{1}{n^{s-1}}\right)\right] &= \overbrace{\sum_{n=1}^{\infty}\frac{1}{n^s}}^{\text{Riemann Zeta}}+ \overbrace{\sum_{n=1}^{\infty}\frac{1}{s-1}\left(\frac{1}{(n+1)^{s-1}} – \frac{1}{n^{s-1}}\right)}^{\text{Telescoping}} \\
&= \zeta(s) + \frac{1}{1 – s}
\end{align*}
Therefore
$$\frac{1}{s-1}+\sum_{n=1}^{\infty}\left[\frac{1}{n^s}+\frac{1}{s-1}\left(\frac{1}{(n+1)^{s-1}} – \frac{1}{n^{s-1}}\right)\right] = \frac{1}{s-1} + \zeta(s) + \frac{1}{1 – s}$$
resulting in the Riemann Zeta function $\zeta(s)$
$$\frac{1}{s-1}+\sum_{n=1}^{\infty}\left[\frac{1}{n^s}+\frac{1}{s-1}\left(\frac{1}{(n+1)^{s-1}} – \frac{1}{n^{s-1}}\right)\right]= \zeta(s)$$
$\blacksquare$
Questions
I'm curious if that's a valid proof? Also, searching for that formula, it was mentioned in these posts below but I couldn't find any reference to its derivation as they're more focused on proving convergence?
Best Answer
We could rewrite the sum as (provided the sum converges): $$\frac1{s-1}+\zeta(s)+\frac1{s-1}(\zeta(s-1)-1-\zeta(s-1))$$Simplifying we get the desired claim. Your proof is correct but it is overkill.