Find the sum explicitly in the interval of convergence of the series.
$\sum _{n=1}^{\infty }\:\frac{n}{n+1}\left(2x-1\right)^n$
Here is the work that I did:
$let\:t\:=\:2x-1$
$\frac{n}{n+1}\:=\:1\:-\:\frac{1}{n+1}$
$\sum _{n=1}^{\infty }\:\frac{n}{n+1}t^n\:=\:\sum _{n=1}^{\infty }\:t^n\:-\:\sum \:_{n=1}^{\infty \:}\:\frac{1}{n+1}t^n\:=\:\frac{1}{1-t}-\sum _{n=1}^{\infty }\:\frac{1}{n+1}t^n$
$f\left(t\right)\:=\:\sum _{n=1}^{\infty }\:\frac{1}{n+1}t^n$
$tf\left(t\right)\:=\:\sum _{n=1}^{\infty }\:\frac{1}{n+1}t^{n+1}$
And here is where my work differs from the solution of the problem.
While I do: $\frac{d}{dx}\left(tf\left(t\right)\right)\:=\:\sum _{n=1}^{\infty }\:t^n\:=\:\frac{1}{1-t}$
The solution says: $\frac{d}{dx}\left(tf\left(t\right)\right)\:=\:\sum _{n=1}^{\infty }\:t^n\:=\:\frac{t}{1-t}$
Following by an integration step and dividing by t. Which one is the correct approach and why?
$\sum _{n=1}^{\infty }\:t^n = t+t^2+\cdots \implies \sum _{n=1}^{\infty }\:t^n= t (1+t+t^2+\cdots)=t + t\sum _{n=1}^{\infty }\:t^n \implies \sum _{n=1}^{\infty }\:t^n = \frac{t}{1-t}$. It seems you misread the sum as $\sum _{n=0}^{\infty }\cdots$