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1

We have, for any $\alpha>1$: $$ \sum_{n\geq 1}\frac{(-1)^n}{n^\alpha} = (2^{1-\alpha}-1)\cdot\zeta(\alpha) \tag{1}$$ hence the main result (or the secondary one up to a shift of the summation variable) follows from considering the opposite of the derivative with respect to $\alpha$ of both sides, followed by an evaluation of the limit for $\alpha\to 1^+$. ...


0

Here is a proof for your expression. Given $$\gamma=\lim_{n \to \infty}\left(H_n-log(n)\right)$$ and $$ \lim_{n \to \infty} log\left( \frac{n^2}{n(n+1)}\right)=0$$ we have $$lim_{n \to \infty}\left(2H_n-H_{n(n+1)}\right) = lim_{n \to \infty}\left(2H_n-H_{n(n+1)}\right)- \lim_{n \to \infty} log\left( \frac{n^2}{n(n+1)}\right)$$ $$=lim_{n \to ...


7

From the continuous fraction expansion, the seventh convergent is $$\gamma \approx \frac{15}{26}$$ From the limit definition $$\begin{align} \gamma &= \lim_{n \to \infty} {\left(2H_n-\frac{1}{6}H_{n^2+n-1}-\frac{5}{6}H_{n^2+n}\right)} \\ &= ...


1

Your analysis is correct. I don't think it really gives insight into Euler's identity per se, but it does help illustrate some geometric intuition about complex arithmetic, which is a great way to understand Euler's identity. First, $r e^{i \theta} = r (\cos \theta + i \sin \theta)$ represents a point in the complex plane in polar coordinates, with a ...



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