Why does ∑(-1)^n/(n+1) at x=1 equal log(2)?

[DottZakapa]

Homework Statement

$\sum_{n=0}^\infty (-1)^n \frac {x^\left(n+1\right)}{n+1}$ for x=1

Relevant Equations

power series

$\sum_{n=0}^\infty (-1)^n \frac {x^\left(n+1\right)}{n+1}$ for x=1

$\sum_{n=0}^\infty (-1)^n \frac {1^\left(n+1\right)}{n+1}$

i've tried leibniz test but i can only find that it converges

why is this power equal to $log(2)$?

i've also tried with $\sum_{n=0}^\infty\log \left (1+\frac 1 {n+1}\right)$

but does not lead to that result

what am i missing ?

 

[vela]

Hint: Try integrating the series for $\frac{1}{1+x}$.

 

[DottZakapa]

Hint: Try integrating the series for $\frac{1}{1+x}$.

which is log(1+x)

ok, didn't understood that if it converges i could plug there the value of x to see at which value converges. Thank you.