Show that $\sum_{n=1}^{\infty} nx^n/(n-1)! = e^xx(x+1)$

Question

Please excuse my relatively novice skills, I'm first year (of 5) on my masters in mathematics. I'm trying to show that $$ \sum_{n=1}^{\infty} \frac{nx^n}{(n-1)!} = e^xx(x+1), \forall x. $$ I already know that the series is convergent for all $x$, but struggling in showing how it evaluates to this. All I can think of, is the Taylor series for $e^x=\sum_{n=0}^{\infty}x^n/n!$, but I can't seem to incorporate this. Any help is appreciated.

Answer 1

$$\begin{aligned}\sum_{n=1}^{\infty}\frac{nx^{n}}{(n-1)!} & =\sum_{n=0}^{\infty}\frac{(n+1)x^{n+1}}{n!}\\ & =\sum_{n=0}^{\infty}\frac{nx^{n+1}}{n!}+x\sum_{n=0}^{\infty}\frac{x^{n}}{n!}\\ & =\sum_{n=1}^{\infty}\frac{nx^{n+1}}{n!}+xe^{x}\\ & =\sum_{n=1}^{\infty}\frac{x^{n+1}}{\left(n-1\right)!}+xe^{x}\\ & =x^{2}\sum_{n=1}^{\infty}\frac{x^{n-1}}{\left(n-1\right)!}+xe^{x}\\ & =x^{2}e^{x}+xe^{x}\\ & =e^{x}x\left(x+1\right) \end{aligned} $$

Answer 2

Since$$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\cdots,$$you know that$$xe^x=x+x^2+\frac{x^3}2+\frac{x^4}{3!}+\cdots$$and that$$x^2e^x=x^2+x^3+\frac{x^4}2+\frac{x^5}{3!}+\cdots$$Therefore,\begin{align}e^xx(x+1)&=xe^x+x^2e^x\\&=x+\left(1+1\right)x^2+\left(\frac12+1\right)x^3+\left(\frac1{3!}+\frac12\right)x^4+\cdots\end{align}Now, note that$$(\forall n\in\mathbb{N}):\frac1{(n-1)!}+\frac1{(n-2)!}=\frac n{(n-1)!}.$$

Answer 3

Hint:

Split the function as

$$x^2e^x+xe^x=\sum_{n=0}^\infty\frac{x^{n+2}}{n!}+\sum_{n=0}^\infty\frac{x^{n+1}}{n!}.$$

Then regroup the terms with the same exponents.

$$\frac1{(n-2)!}+\frac1{(n-1)!}=\frac{n-1+1}{(n-1)!}=\frac{n}{(n-1)!}.$$

Answer 4

An other way to prove it :

We have :

$$\forall x \in \mathbb{R}, \sum_{n = 0}^{\infty} \frac{x^n}{n!} = e^x$$ Hence we have : $$\forall x \in \mathbb{R}, \sum_{n = 1}^{\infty} \frac{x^{n-1}}{(n-1)!} = (e^x)' = e^x$$

It follows that :

$$\forall x \in \mathbb{R}, \sum_{n = 1}^{\infty} \frac{x^n}{(n-1)!} = xe^x$$

And we have :

$$\forall x \in \mathbb{R}, \sum_{n = 1}^{\infty} \frac{nx^{n-1}}{(n-1)!} = (xe^x)' = e^x(x+1)$$

Hence we get the desired result :

$$\forall x \in \mathbb{R}, \sum_{n = 1}^{\infty} \frac{nx^n}{(n-1)!} = e^xx(x+1)$$