How can one show that $\sum_{n=0}^\infty\frac{n}{n!}=e$?

Question

How can one show that $\sum_{n=0}^\infty\frac{n}{n!}=e$?

I understand that $$ \sum_{n=0}^\infty\frac{x^n}{n!}=e^x $$ and that letting $x=1 $would give $$ \sum_{n=0}^\infty\frac{1}{n!}=e $$

But why does the sum $\sum_{n=0}^\infty\frac{n}{n!}$ give an answer of $e$ also?

Answer 1

$$\begin{array} {rcl} % \displaystyle e^x & = & \displaystyle \sum_{n = 0}^{\infty} \frac{x^n}{n!} \\ % \displaystyle\frac{d}{dx}e^x & = & \displaystyle \frac{d}{dx}\sum_{n = 0}^{\infty} \frac{x^n}{n!} \\ % e^x & = & \displaystyle \sum_{n = 0}^{\infty} \frac{nx^{n-1}}{n!} \\ % e^1 & = & \displaystyle \sum_{n = 0}^{\infty} \frac{n1^{n-1}}{n!} \\ \end{array}$$

Answer 2

Suppose $e^x=a_0+a_1x+a_2x^2+a_3x^3+...$ try to find the constants by evaluating $e^0$ and then differentiating to eliminate the current $0$th degree term.

Here are the first three terms : $$e^0=a_0+a_1*0+...=1\Leftrightarrow a_0=1$$ $$\frac{d}{dx}(e^x)=e^x=\frac{d}{dx}(a_0+a_1x+a_2x^2+...)=a_1+2a_2x+3a_3x^2+...$$ $$e^0=a_1+2a_2*0+3a_3*0^2+...=1\Leftrightarrow a_1=1$$ $$\frac{d}{dx}(e^x)=e^x=\frac{d}{dx}(a_1+2a_2x+3a_3x^2+...)=2a_2+(3*2)a_3x+...$$ $$e^0=2a_2+(3*2)a_3*0\,+...=1\Leftrightarrow a_1=\frac{1}{2}$$ You'll find that $e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$ switch $1$ for $x$ and you get your result (refer to The Count's answer).

Answer 3

Differentiate the series $e^x=\sum_{n=0}^{\infty}\dfrac {x^n}{n!}$ term by term. Get $\sum_{n=0}^{\infty}\dfrac {nx^{n-1}}{n!}$. Then set $x=1$. Note that $(e^x)'=e^x$ (for$\sum_{n=0}^{\infty}\dfrac {nx^{n-1}}{n!}=\sum_{n=0}^{\infty}\dfrac {x^n}{n!}$) .

Answer 4

Note that $n/n!=1/(n-1)!$, which is not meaningfully distinct in the context of the infinite sum from the terms $1/n!$.

Answer 5

In fact $$ \sum_{n\geq 0} \frac{n}{n!}\stackrel{(1)}{=}\sum_{n\geq 1} \frac{n}{n!}= \sum_{n\geq 1} \frac{1}{(n-1)!}\stackrel{(2)}{=}\sum_{n\geq 0} \frac{1}{n!}=e^1=e $$ (1) comes from the fact that the first term is zero, (2) is a shift of indices by one.

You can repeat this shifting procedure ad infinitum and obtain, for each fixed $k$ $$ \sum_{n\geq 0} \frac{n(n-1)\ldots (n-k)}{n!}=e $$ because your sum 'starts' at index $k+1$.