Given the power series, using differentiation and integration, find the function represented by the power series:
$$S(x)=\sum_{n=1}^\infty nx^{n-1}$$
By integrating this I get $\sum_{n=0}^\infty x^n$ and I know that this is the function $f(x)=\frac1{1-x}$, but I am unaware if I am missing any steps or simply stating the function right after integrating is correct?
Best Answer
You don't even need differentiation or integration. $S(x)$ is well defined in the unit ball and: $$ (1-x)\cdot S(x) = (1-x)\sum_{n=1}^{+\infty} n x^{n-1} = \sum_{n\geq 1}\left(n x^{n-1}-n x^{n}\right)=\sum_{n\geq 1}x^{n-1} = \frac{1}{1-x}$$ from which: $$ S(x) = \frac{1}{(1-x)^2} $$ follows. As an alternative: $$ S(x) = \frac{d}{dx}\sum_{n\geq 1}x^n = \frac{d}{dx}\left(\frac{1}{1-x}-1\right)=\frac{1}{(1-x)^2}.$$