I treid to show that $$\sum _{n=1}^{\infty} \frac{\left(-1\right)^{n-1}}{n}x^n = \log(1+x)$$ for $\mid x\mid <1$ by showing that $$\sum _{n=1}^{\infty} (-1)^{n-1} x^{n-1} = \frac {1}{x+1}$$ uniformly and then integrate both sides of equation due to uniform convergence. But we know that $\sum _{n=1}^{\infty} x^{n-1}$ doesn't converge uniformly for $\mid x\mid <1$ so I can not use Abel's test.
Thanks for any help to show $\sum _{n=1}^{\infty} \frac{\left(-1\right)^{n-1}}{n}x^n = \log(1+x)$ for $\mid x\mid <1$.
Take $x\in(-1,1)$ and take $r\in(|x|,1)$. Then $\sum_{n=1}^{\infty}(-1)^{n-1}x^{n-1}$ converges uniformly on $[-r,r]$ to $\frac1{x+1}$. And so$$(\forall t\in[-r,r]):\sum_{n=1}^\infty\int_0^t(-1)^{n-1}x^{n-1}\,\mathrm dx=\log(t+1).$$In particular, $\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n-1}}nx^n=\log(x+1)$.