Limit with zeta and Gamma function

Let $\zeta$ denote the Riemann zeta function and $\Gamma$ denote the Euler’s Gamma function. Prove that

$\frac{1}{2} \lim_{n \rightarrow +\infty} \left [ \zeta\left ( 1 + \frac{1}{n} \right ) - \Gamma \left ( \frac{1}{n} \right ) \right ] = \gamma$

where $\gamma$ stands for the Euler – Mascheroni constant.

Solution

Using the Laurent expansion of the $\zeta$ function we have that

$\begin{align*} \zeta(s) \sim \frac{1}{s-1} + \gamma + \mathcal{O} \left ( s-1 \right ) &\Rightarrow \zeta\left ( 1 + \frac{1}{n} \right ) \sim \frac{1}{1+\frac{1}{n} - 1} + \gamma + \\ & \quad \quad \quad \quad \quad \quad \quad +\mathcal{O} \left ( 1 + \frac{1}{n} - 1 \right )\\ &\Rightarrow \zeta \left ( 1 + \frac{1}{n} \right ) \sim n + \gamma + \mathcal{O} \left ( \frac{1}{n} \right) \end{align*}$

Using the Laurent expansion of the $\Gamma$ function we have that

$\begin{align*} \Gamma(s) \sim \frac{1}{s} - \gamma + \mathcal{O}(s) &\Rightarrow \Gamma \left ( \frac{1}{n} \right ) \sim n - \gamma + \mathcal{O} \left ( \frac{1}{n} \right ) \end{align*}$

The result now follows.

Limit with zeta and Gamma function

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