Taylor series for $\log(1+x)$ and its convergence

Question

I know the expression for the Taylor series of $\log(1+x)$.

However, I don't understand how to prove the convergence of the series for $x>1$ and its divergence for $x<1$. Can someone explain the reason for this?

Answer 1

We have (i) convergence if $|x|\lt 1$, and divergence if $|x|\gt 1$. This can be done by using the Ratio Test.

We also have (ii) convergence at $x=1$ and divergence at $x=-1$. For $x=1$, we have an alternating series. For $x=-1$, we get a close relative of a familiar divergent series.

Answer 2

For $0 \le r < 1$ $$\frac{1}{1-r} = \sum_{k=0}^\infty r^k$$ Integrate both pats $ln(1-r) = \sum_{k=1}^\infty \frac{r^{k}}{k} $

For $-1 < r \le 0$ $$\frac{1}{1+r} = \sum_{k=0}^\infty (-1)^kr^k$$ Integrate both pats $ln(1+r) = \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}r^{k} $