Proof of convergence for
\( \displaystyle{ \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} } \)
for Re(s) > 1:
I will use that definition and not analytical continuations of that definition for Re(s) <= 1.
First off, ζ(s) > 0 for all s, supplying a lower bound. So we must look for an upper bound. Since this series diverges for Re(s) <= 0, we look at Re(s) > 0. Finding a lower bound for each n thus gives an upper bound for the series. We can also find a lower bound by finding an upper bound for each n, something that will improve our convergence estimate.
Lower bound for each n: (1) >= 1, (2, 3) >= 2, (4, 5, 6, 7) >= 4, ... giving 2
k of n = 2
k
Upper bound for each n: (1) <= 1, (2) <= 2, (3, 4) <= 4, (5, 6, 7, 8) <= 8, ... giving 1, then 2
k of each n = 2
k+1
Thus,
\( \displaystyle{ 1 + \sum_{k=0}^\infty \frac{2^k}{2^{s(k+1)}} < \zeta(s) < \sum_{k=0}^\infty \frac{2^k}{2^{sk}} } \)
Both bounds are geometric series:
\( \displaystyle{ 1 + \sum_{k=0}^\infty \frac{2^k}{2^{s(k+1)}} < \zeta(s) < \sum_{k=0}^\infty \frac{2^k}{2^{sk}} } \)
reducing to
\( \displaystyle{ 1 + 2^{-s} \sum_{k=0}^\infty 2^{-(s-1)k} < \zeta(s) < \sum_{k=0}^\infty 2^{-(s-1)k}} \)
This means that the series form of ζ(s) converges for Re(s) > 1 and diverges for Re(s) <= 1.
There's a lot more at
Riemann zeta function
