# Order of growth of real $x_{n}$ such that $\zeta(x_{n}) = 1 + 1/2^{n}$

On a lark, I decided to calculate (via Newton's method and using mpmath) the real $x_{n}$ such that $\zeta(x_{n}) = 1 + 1/2^{n}$ for as many $n\in\mathbb{N}$ as I could. What sort of surprised me is that as $n$ grew, $x_{n}$ started getting very close to $n$.

For $n=30$, $x_{30}\approx 30.0000075250944636973647\ldots$.
For $n=100$, $x_{100}\approx 100.00000000000000000354853124\ldots$

Is it true that $\zeta(2m)=(-1)^{m+1}\frac{B_{2m}(2\pi)^{2m}}{2(2m)!}$ grows as $1+\frac{1}{2^{2m}}$?

(here is a link to a file containing $x_{n}$ for small $n$ to many digits of accuracy)

EDIT: as @sos440 pointed out, the series definition of the zeta function (that is valid on the real $s>1$) is $\sum_{n=1}^{\infty} \frac{1}{n^{s}}$, and taking the first two terms one obtains: $1 + \frac{1}{2^{n}} + \sum_{n=3}^{\infty} \frac{1}{n^{s}}$.

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Just carefully look at the defining series of the Riemann zeat function. Then you will soon find a reasonable explanation to this phenomenon. –  sos440 Jun 15 '11 at 10:11
@sos440: oh, I should have seen that. –  deoxygerbe Jun 15 '11 at 10:30