Show $$\ln(2)=\sum_{i=1}^{\infty}\frac{(-1)^{n+1}}{n}$$
We know que if $|z|<1$.
$$\ln(1-z)=-\sum_{i=1}^{\infty}\frac{(z)^{n}}{n}$$
But $z=-1$ for our case and I wouldn't meet the condition $|z|<1$.
Asked
Active
Viewed 55 times
0
-
Also: https://math.stackexchange.com/q/146471/42969, https://math.stackexchange.com/q/2086393/42969 – all found with Approach0 – Martin R Jul 13 '19 at 19:16
-
"$z=-1$...wouldn't meet the condition $|z|<1$". It does not have to. All that is needed is convergence, which could happen at some (though not all) unit-sized values of $z$. – Oscar Lanzi Jul 13 '19 at 19:20