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

asmaier
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2 Answers2

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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 $x \to \infty$.

Community
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Olivier Oloa
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$\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

marty cohen
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