We have the series $\sum_{n=0}^\infty \frac{n}{n+1}x^n$.
Give a function $f(x)$such that this function is equal to this series for each $x\in \Bbb R$ for which the series converges.
I've calculated the radius of convergence, which was 1. So $-1 < x<1$. And when you write out the series you get
$\sum_{n=0}^\infty \frac{n}{n+1}x^n$=$0+\frac{x}{2}+\frac{2x^2}{3}...$
I don't see how a good $f(x)$ for this series. I'd appreciate some hints about what kind of function I'm looking for.
$$f(x)=\sum_{n=0}^\infty \frac{n}{n+1}x^n=\sum_{n=0}^\infty \frac{n+1-1}{n+1}x^n=\sum_{n=0}^\infty x^n-\sum_{n=0}^\infty \frac{x^n}{n+1}$$
$$\ln (1-x)= -\sum_{n=\color{blue}1}^\infty \frac{x^n}{n} \implies \sum_{n=\color{blue}0}^\infty \frac{x^n}{n+1}=\frac{-\ln (1-x)}{x}$$
$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n $$
$$\text{Therefore}: ~f(x)= \frac{1}{1-x}+\frac{\ln (1-x)}{x}$$
$$ \sum_{n=0}^{\infty} \frac{n+1-1}{n+1} x^{n}=\sum_{n=0}^{\infty} x^{n}-\sum_{n=0}^{\infty} \frac{x^{n}}{n+1} $$
is only allowed if you have shown that $\sum_{n=0}^{\infty} x^{n}$ is conv. with a sum $A$ and $\sum_{n=0}^{\infty} \frac{x^{n}}{n+1}$ is conv. with a sum $B$.
– Sorry May 29 '20 at 16:31$$\sum_{n=0}^\infty \frac{n}{n+1}x^n=\sum_{n=0}^\infty x^n-\dfrac1x\sum_{n=0}^\infty\dfrac{x^{n+1}}{n+1}$$
Now $\ln(1-y)=-\sum_{u=1}^\infty\dfrac{y^u}u$
Now see Taylor series for $\log(1+x)$ and its convergence and this
