Finding a function $f$ such that a given series is its Taylor series

Question

Given the following series $$\sum_{n=1}^\infty (n+1)x^n$$ find a function $f$ such that this series is the Taylor series of $f$.

So far I have:

$$\sum_{n=1}^\infty (n+1)x^n= \sum_{n=1}^\infty nx^n +\sum_{n=1}^\infty \frac{x^n}{n} = \frac{1}{1-x}\sum_{n=1}^\infty n +\sum_{n=1}^\infty \frac{x^n}{n}$$

My first problem was that $\sum_{n=1}^\infty n$ diverges. The second problem was the last sum $$\sum_{n=1}^\infty \frac{x^n}{n}$$ I recognized $\ln(1+x)$, but then you would need to have

$$\sum_{n=1}^\infty (-1)^{n-1}\frac{x^n}{n}$$

I think you can probably manipulate the sum such that you get the $\ln(1+x)$ but I'm not sure how to do that.

Answer 1

Hint: $$\sum_{n=1}^\infty (n+1)x^n= \sum_{n=1}^\infty(x^{n+1})'=\Bigl(\sum_{n=2}^\infty x^{n}\Bigr)'.$$

Answer 2

hint:

$$\sum_{n=1}^{\infty}nx^n=\sum_{n=1}^{\infty}x\frac{d}{dx}x^n$$

be careful on the bounds on $x$