How to prove that $\sum _{n=0}^{\infty }\:\frac{(x^n)'}{(n-1)!} = e^{x}(x-1)$

Question

I am trying to prove that $$\sum _{n=0}^{\infty }\:\frac{\left(x^n\right)'}{\left(n-1\right)!} = e^{x}(x+1)\tag 1$$

---

This sum is very similar to the derivative of exponential $(e^x)' = \sum _{n=0}^{\infty } \frac{(x^{n})'}{n!} = \sum_{n=0}^{\infty} \frac{x^{n-1}}{(n-1)!}$ but there is that $n$ factor that I can't get rid off.

Any ideas?

Answer 1

We have \begin{align} \sum_{n=1}^{\infty} \frac{nx^{n-1}}{(n-1)!} &= \sum_{n=1}^{\infty} \frac{(n-1+1)x^{n-1}}{(n-1)!}\\ &= \sum_{n=1}^{\infty} \frac{(n-1)x^{n-1}}{(n-1)!} + \sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!}\\ &= x\sum_{n=2}^{\infty} \frac{x^{n-2}}{(n-2)!} + e^x\\ &= e^x(x+1). \end{align}

Answer 2

Hint: $x^n=x\cdot x^{n-1}$ the product rule says that $(x^n)'=(x\cdot x^{n-1})'=x^{n-1}+(n-1)x\cdot x^{n-2}.$

Answer 3

An alternative version.

$$\int_0^x\sum_{n=1}^\infty \frac{(x^n)'}{(n-1)!}dx = \sum_{n=1}^\infty \frac{x^n}{(n-1)!} = x \sum_{n-1}^\infty \frac{x^{n-1}}{(n-1)!} = x \cdot e^x$$

Derivate the result and you get

$$(x e^x)' = e^x+xe^x = (1+x)e^x$$