Analytic Continuation of Zeta Function using Bernoulli Numbers In my complex analysis textbook by Stein and Shakarchi, as an exercise, I am supposed to extend $\zeta(s)$ to the entire complex plane using Bernoulli numbers, but I am stuck.
I can prove that
$$ \zeta(s) = \frac{1}{\Gamma(s)}  \int_0^1 \frac{x^{s-1}}{e^{x}-1} dx + \frac{1}{\Gamma(s)}  \int_1^\infty \frac{x^{s-1}}{e^{x}-1} dx $$
I can further prove that the second integral is an entire function and that given the generating function for the Bernoulli Numbers:
$$ \frac{x}{e^x -1} = \sum_{m=0}^\infty \frac{B_m}{m!} x^m $$
that 
$$ \int_0^1 \frac{x^{s-1}}{e^x-1}dx = \sum_{m=0}^\infty \frac{B_m}{m!(s+m-1)} $$
It is clear that there is a pole at $s=1$, but beyond that I don't have any idea why the last summation converges for all $s \neq 1$, which would prove the analytic continuation of $\zeta(s)$ to the entire complex plane.
This is a problem in a first year graduate student textbook, so I imagine that there is a short solution.  Thanks for the help.
 A: Since $\frac{z}{e^z-1}$ is holomorphic on the disk $\{ z : \lvert z\rvert < 2\pi\}$, and has poles in $\pm 2\pi i$, the radius of convergence of the series
$$\sum_{m=0}^\infty \frac{B_m}{m!}z^m$$
is $2\pi$. Hence, by the Cauchy-Hadamard formula, 
$$\limsup_{m\to\infty} \left(\frac{\lvert B_m\rvert}{m!}\right)^{1/m} = \frac{1}{2\pi},$$
so for every $R < 2\pi$, for all large enough $m$, we have
$$\frac{\lvert B_m\rvert}{m!} < \frac{1}{R^m}.$$
We know a bit more from the representations of the cotangent: for $\mu\geqslant 1$ we have $B_{2\mu+1} = 0$, and
$$\frac{B_{2\mu}}{(2\mu)!} = \frac{2(-1)^{\mu+1}}{(2\pi)^{2\mu}}\zeta(2\mu),$$
so for all $m\geqslant 0$
$$\frac{\lvert B_m\rvert}{m!} < \frac{4}{(2\pi)^m}.$$
Since the series converges uniformly on the interval of integration, we can interchange integration and summation, and
$$\int_0^1 \frac{x^{s-1}}{e^x-1}\,dx = \int_0^1 \sum_{m=0}^\infty \frac{B_m}{m!} x^{m+s-2}\,dx = \sum_{m=0}^\infty \frac{B_m}{m!}\int_0^1 x^{m+s-2}\,dx = \sum_{m=0}^\infty \frac{B_m}{m!(s+m-1)}.$$
For any $K > 0$, if $\lvert s\rvert \leqslant K$ and $m \geqslant K+2$, we have $\lvert s+m-1\rvert \geqslant 1$, so the sum
$$\sum_{m = \lceil K\rceil+2}^\infty \frac{B_m}{m!(s+m-1)}$$
converges uniformly on the disk $\{ s : \lvert s\rvert \leqslant K\}$ by the Weierstraß $M$-test. Hence
$$H(s) = \sum_{m=0}^\infty \frac{B_m}{m!(s+m-1)}$$
is an entire meromorphic function with simple poles in $1-m,\, m\in\mathbb{N}$.
Since $\Gamma(s)$ has simple poles in $-k,\,k\in\mathbb{N}$, and no zeros, the function
$$\frac{H(s)}{\Gamma(s)} = \frac{1}{\Gamma(s)}\sum_{m=0}^\infty \frac{B_m}{m!(s+m-1)}$$
is an entire meromorphic function with only one simple pole, in $1$, with resdiue $\frac{B_0}{0!\Gamma(1)} = 1$.
