When I deal with an estimation in analysis,I hope the equality given below is true.The geometry series on the numerator and the estimates $\sum_{n=1}^N {1\over n}\geqslant clogN$ we all know indulge me to believe my estimation is right.However,I can't link these two things to verify my estimation. $$\sum_{n=1}^\infty {r^n\over n}\geqslant log({1\over 1-r})$$ If it's true,can anyone gives me the proof of the equality?If it is not true,I'll try scaling the left side more exactly.
($0\lt r \lt 1$)
Let $$f(r)=\sum_{n=1}^\infty\frac{r^n}n$$ (where $|r|<1$). Then $f(0)=0$ and $$f'(r)=\sum_{n=1}^\infty\frac{nr^{n-1}}n=\sum_{n=0}^\infty r^n=\frac1{1-r}.$$ Now consider $g(r):=\ln\frac1{1-r}=-\ln (1-r)$. We have $g(0)=0$ and $g'(r)=\frac1{1-r}$. We conclude that $f-g$ os constant on $(-1,2)$ and that constant is $0$. In other words $$\sum_{n=1}^\infty\frac{r^n}n=\ln\frac1{1-r}$$ for $|r|<1$.