# 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 $$\lim_{x \to \infty} \frac{1}{2}\left(\zeta(1+\frac{1}{x}) - \Gamma(\frac{1}{x})\right) = \gamma$$

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 $x \to \infty$.
$\lim_{x \to \infty}\sum_{n=1}^{\infty} \frac{1}{n^{1+1/x}} - x = \gamma$
$\begin{array}\\ \sum_{n=1}^{m} \frac{1}{n^{1+1/x}} &=\sum_{n=1}^{m} \frac{1}{n}\frac{1}{n^{1/x}}\\ &=\sum_{n=1}^{m} \frac{1}{n}e^{-\ln n/x}\\ &\approx \sum_{n=1}^{m} \frac{1}{n}(1-\ln n/x) \qquad\text{for large } x\\ &= \sum_{n=1}^{m} \frac{1}{n}-\frac1{x}\sum_{n=1}^{m}\frac{\ln n}{n}\\ &= \ln(m)+\gamma+o(1)-\frac1{x}\sum_{n=1}^{m}\frac{\ln n}{n}\\ &= \ln(m)+\gamma+o(1)-\frac1{x}(\frac{\ln^2(m)}{2}+O(1)) \qquad(*)\\ &= \ln(m)+\gamma+o(1)-\frac{\ln^2(m)}{2x}+O(\frac1{x})\\ &= \ln(m)+\gamma+o(1)-\frac{\ln^2(m)}{2x}+O(\frac1{x})\\ \end{array}$
Therefore $\sum_{n=1}^{m} \frac{1}{n^{1+1/x}} -\ln(m)-\gamma =o(1)-\frac{\ln^2(m)}{2x}+O(\frac1{x})$