Zeta Function $\zeta(1\pm1/n)$ and Euler's constant.

Question

How do I show that $$\lim_{n\to\infty}{\zeta(1+1/n)+\zeta(1-1/n)}=2\gamma$$ and $$\lim_{n\to\infty}{\zeta(1+1/n)-\zeta(1-1/n)}=\infty,$$ where $\gamma$ is the Euler's constant?

Answer 1

$$ \zeta(s) = \frac{1}{s-1} + \gamma + 0.07281584548367672486... \cdot (s-1) + \mbox{more} $$ and $$ \zeta(s) - \frac{1}{s-1} $$ is holomorphic and entire, meaning that the radius of convergence of the indicated series around $1$ is $\infty.$

see http://en.wikipedia.org/wiki/Stieltjes_constants

Answer 2

For small values of $x$, Taylor expansions give $$\zeta(1+x)=\frac{1}{x}+\gamma -\gamma _1 x+\frac{\gamma _2 x^2}{2}-\frac{\gamma _3 x^3}{6}+O\left(x^4\right)$$ $$\zeta(1-x)=-\frac{1}{x}+\gamma +\gamma _1 x+\frac{\gamma _2 x^2}{2}+\frac{\gamma _3 x^3}{6}+O\left(x^4\right)$$ where appear the Euler constant and a few of the Stieltjes constants.

For the problem you posted, you can limit yourself to the first two terms.

I am sure that you can take from here.