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

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?

• In the picture on the bottom sum the numerator should be n, apologies.
– user677704
Jun 1, 2019 at 2:51
• Are you familiar with derivatives ? Jun 1, 2019 at 2:53
• Your question is rather unclear. Do you want to prove: $\sum _{n=0}^{\infty }\:\frac{x}{n!}=e?$ Jun 1, 2019 at 3:12
• ‘No chance’ yes that was my question more or less, I just didn’t understand why different sums all add to e
– user677704
Jun 1, 2019 at 3:13
• It can be proven that $\sum _{n=0}^{\infty }\:\frac{x^n}{n!}=e^x$. However, in your expression, you don't raise x to the power n. Jun 1, 2019 at 3:15

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

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 $$k+1$$.

• This is what I meant in my answer. Nice presentation. +1 Jun 1, 2019 at 3:14
• So I, in fact, unfolded your idea ? (+1) Jun 1, 2019 at 3:15
• Teamwork! I like it. Jun 1, 2019 at 3:16

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

• Thanks for your response it’s really clear and perfectly understood, I appreciate it. So would I be right in assuming you can repeat this process ad infinitum and acquire many different expressions which all sum to e?
– user677704
Jun 1, 2019 at 3:15
• Yes, in this case the multipliers will be $$n,n(n-1),n(n-1)(n-2),\ldots$$ and so on ... Jun 1, 2019 at 3:18
• That’s incredibly interesting, aside from showing this with basic differentiation would you know where to find a geometric or more intuitive proof. The logic is perfectly reasonable but I prefer to see why.
– user677704
Jun 1, 2019 at 3:25
• @User3457884334 I added repetition of this process ad infinitum in my answer. Jun 1, 2019 at 3:26
• I understand that but wondered if there was another proof to show it geometrically.
– user677704
Jun 1, 2019 at 3:28

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).

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$$\sum_{n=0}^{\infty}\dfrac {nx^{n-1}}{n!}=\sum_{n=0}^{\infty}\dfrac {x^n}{n!}$$) .