Interesting formula for Euler-Mascheroni constant? Playing around with Wolfram Alpha I found a possible formula for the Euler-Mascheroni constant $\gamma$:
$$
\lim_{x \to \infty}\sum_{n=1}^{\infty} \frac{1}{n^{1+1/x}} - x = \gamma
$$
This can also be written in terms of the Riemann Zeta function like
$$
\lim_{x \to \infty}\zeta(1+\frac{1}{x}) - x = \gamma
$$
I find this remarkable, because this seems to imply that $\zeta(1)$ is "a little bit bigger than infinity"
$$
\infty + \gamma = \zeta(1)
$$
Is this really a valid formula for $\gamma$ ? 
Update: Curiously I also found a similar formula for the Gamma function
$$
\lim_{x \to \infty}\Gamma(\frac{1}{x}) - x = -\gamma
$$
Combining both equations generates again an interesting formula
$$
\lim_{x \to \infty} \frac{1}{2}\left(\zeta(1+\frac{1}{x}) - \Gamma(\frac{1}{x})\right) = \gamma
$$
 A: Your observation comes from the Laurent series expansion of the Riemann zeta function near $1$ (see here) giving, as $t \to 0^+$,
$$
\zeta(1+t) = \frac{1}{t} + \gamma + o(1)
$$ then just set $t=\dfrac1x$ with $x \to \infty$.
A: $\lim_{x \to \infty}\sum_{n=1}^{\infty} \frac{1}{n^{1+1/x}} - x = \gamma
$
Let's look at the partial sums.
$\begin{array}\\
\sum_{n=1}^{m} \frac{1}{n^{1+1/x}}
&=\sum_{n=1}^{m} \frac{1}{n}\frac{1}{n^{1/x}}\\
&=\sum_{n=1}^{m} \frac{1}{n}e^{-\ln n/x}\\
&\approx \sum_{n=1}^{m} \frac{1}{n}(1-\ln n/x)
\qquad\text{for large } x\\
&= \sum_{n=1}^{m} \frac{1}{n}-\frac1{x}\sum_{n=1}^{m}\frac{\ln n}{n}\\
&= \ln(m)+\gamma+o(1)-\frac1{x}\sum_{n=1}^{m}\frac{\ln n}{n}\\
&= \ln(m)+\gamma+o(1)-\frac1{x}(\frac{\ln^2(m)}{2}+O(1))
\qquad(*)\\
&= \ln(m)+\gamma+o(1)-\frac{\ln^2(m)}{2x}+O(\frac1{x})\\
&= \ln(m)+\gamma+o(1)-\frac{\ln^2(m)}{2x}+O(\frac1{x})\\
\end{array}
$
Therefore
$\sum_{n=1}^{m} \frac{1}{n^{1+1/x}}
-\ln(m)-\gamma
=o(1)-\frac{\ln^2(m)}{2x}+O(\frac1{x})
$
(*) http://www.maths.lancs.ac.uk/~jameson/emnotes.pdf,
p. 14
