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I want to prove that $ \lim \limits_{s \to 0} (\zeta(1+s)+\zeta(1-s))= 2\gamma$ , I divide it to two limits, $\lim \limits_{s \to 1^{+}} (\zeta(s)-\frac{1}{s-1}) = \gamma$ which I proved using the definition for $\gamma = H(\infty)-\ln (\infty)$ and $\zeta(s) = \sum \limits_{n=1}^{\infty} n^{-s} $ for $s>1$ but the definition of $\zeta(s)$ fails when evaluating $\lim \limits_{s\to 1^{-}} (\zeta(s)-\frac{1}{s-1})$, so I want to use $\zeta(s) = \frac{1}{1-2^{-s}} \sum \limits_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^s}$ for $s>0$, the problem for me now is that this new definition does not easily transform to the definition of $\gamma = \lim \limits_{n \to \infty } (H_n -\ln n)$, I want to see a proof of the case when $s\to 1^{-}$ ?

Thanks

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2 Answers 2

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The expression in the OP for the Riemann Zeta function should read

$$\zeta(s)=\frac1{1-2^{\color{red}1-s}}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s}\tag1$$


First, we expand the term $\left(1-2^{1-s}\right)^{-1}$ in $(1)$ at $s=1$ to find that

$$\begin{align} \frac1{1-2^{\color{red}1-s}}&=\frac1{1-e^{(1-s)\log(2)}}\\\\ &=\frac1{\log(2)(s-1)-\frac12 \log^2(2)(s-1)^2+O((s-1)^3)}\\\\ &=\frac1{\log(2)(s-1)}\left(1+\frac{\log(2)(s-1)}2+O(s-1)^2\right)\tag2 \end{align}$$


Next, we expand the series $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^s}$ in $(1)$ at $s=1$ to find that

$$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}=\log(2)-\sum_{n=1}^\infty \frac{(-1)^{n-1}\log(n)}{n}(s-1)+O(s-1)^2\tag3$$


Using $(2)$ and $(3)$ in $(1)$ reveals

$$\zeta(s)-\frac1{s-1}=\frac12 \log(2)-\frac1{\log(2)}\sum_{n=1}^ \infty \frac{(-1)^{n-1}\log(n)}{n}+O(s-1)\tag4$$


It remains to show that the series in $(4)$ in equal to $\frac12\log^2(2)-\gamma \log(2)$. To do so, we apply the Euler-Maclaurin Summation Formula to the sum $\sum_{n=N+1}^{2N} \frac{\log(n)}{n}$. Proceeding, we see that

$$\begin{align} \sum_{n=1}^{2N} \frac{(-1)^{n-1}\log(n)}{n}&=\sum_{n=1}^{2N}\frac{\log(n)}{n}-\sum_{n=1}^N \frac{\log(2n)}{n}\\\\ &=\color{red}{\sum_{n=N+1}^{2N}\frac{\log(n)}{n}}-\color{blue}{\log(2)\underbrace{\sum_{n=1}^N \frac1n}_{=H_N}}\\\\ &=\color{red}{\frac12 \log^2(2N)-\frac12\log^2(N)+O\left(\frac{\log(N)}{N}\right)}\\\\ &-\color{blue}{\log(2)\left(\log(N)+\gamma+O\left(\frac1N\right)\right)}\\\\ &=\frac12\log^2(2)+\log(2)\log(N)+O\left(\frac{\log(N)}{N}\right)\\\\ &-\log(2)\left(\log(N)+\gamma+O\left(\frac1N\right)\right)\\\\ &=\frac12\log^2(2)-\log(2) \gamma+o(1) \end{align}$$

Letting $N\to \infty$, we obtain the result

$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}\log(n)}{n}=\frac12\log^2(2)-\log(2)\gamma\tag5$$


Using $(5)$ in $(4)$ and letting $s\to 1$ yields

$$\bbox[5px,border:2px solid #C0A000]{\lim_{s\to 1}\left(\zeta(s)-\frac1{s-1}\right)=\gamma}$$

as was to be shown!

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We can express both situations using Euler-Maclaurin formula:

$$ \sum_{n=1}^N\frac1n=\log N+\frac12+{1\over2N}-\int_1^N{\overline B_1(x)\over x^2}\mathrm dx $$

Using the definition of Euler-Mascheroni constant, we can see that

$$ \gamma=\frac12-\int_1^\infty{\overline B_1(x)\over x^2}\mathrm dx $$

Now, let's apply Euler-Maclaurin formula to $\zeta(s)$

$$ \zeta(s)=\sum_{n=1}^\infty{1\over n^s}={1\over s-1}+\frac12-s\int_1^\infty{\overline B_1(x)\over x^{s+1}}\mathrm dx $$

Since $|\overline B_1(x)|$ is bounded, we see that the right-most integral converges for $\Re(s)>1$, so the RHS is an analytic continuation. Consequently, we have

$$ \lim_{s\to1}\left\{\zeta(s)-{1\over s-1}\right\}=\frac12-\int_1^\infty{\overline B_1(x)\over x^2}\mathrm dx=\gamma $$

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