Nessun numero è un’isola

$$\begin{array}{rl} \ln(2)=\ln\left(1+1\right)& =\dfrac1{1}-\dfrac1{2}+\dfrac1{3}-\dfrac1{4}+\ldots\\ \ln(3)-\ln(2) =\ln\left(1+\dfrac12\right) & =\dfrac12-\dfrac12\dfrac1{2^2}+\dfrac13\dfrac1{2^3}-\dfrac14\dfrac1{2^4}\ldots \\ \ln(4)-\ln(3) =\ln\left(1+\dfrac13\right)  & =\dfrac13-\dfrac1{2}\dfrac1{3^2}+\dfrac13\dfrac1{3^3}-\dfrac14\dfrac1{3^4}+\ldots \\ & \vdots  \\ \ln(n+1)-\ln(n) =\ln\left(1+\dfrac1n\right)  & =\dfrac1n-\dfrac1{2}\dfrac1{n^2}+\dfrac13\dfrac1{n^3}-\dfrac14\dfrac1{n^4}+\ldots \end{array}$$sommando membro a membro e riordinando otteniamo

$$\begin{array}{rl}  \ln(n+1) = &1+\dfrac12+\dfrac13+\ldots+\dfrac1n+ \\ & -\dfrac12\left(1 + \dfrac1{2^2}+\dfrac1{3^2}+\ldots+\dfrac1{n^2} \right) + \\ & +\dfrac13\left(1+\dfrac1{2^3}+\dfrac1{3^3}+\ldots+\dfrac1{n^3} \right) + \\ & -\dfrac14\left(1+\dfrac1{2^4}+\dfrac1{3^4}+\ldots+\dfrac1{n^4}\right) + \\ & \vdots  \end{array}$$

Spostando $1+\dfrac12+\dfrac13+\ldots+\dfrac1n$ a sinistra e facendo tendere $n$ a infinito otteniamo

$$-\,\gamma=-\dfrac12\zeta(2)+\dfrac13\zeta(3)-\dfrac14\zeta(4)+\ldots$$ovvero$$\gamma=\sum_{n=2}^\infty\ (-1)^n\dfrac{\zeta(n)}n$$

La convergenza è molto lenta dato che $\zeta(n)\simeq1+O\left(\dfrac1{2^n}\right)$. Per ottenere $\gamma$ con una convergenza più veloce si può usare la formula equivalente $$\gamma=1-\ln2+\sum_{n=2}^\infty\ (-1)^n\ \dfrac{\zeta(n)-1}{n}\ .$$