Can we find $ \sum_{n=1}^{\infty}\frac{1+2+\cdots +n}{n!} $? Consider the sequence $$ a_{n} = \sum_{r=1}^{n}\frac{1+2+\cdots +r}{r!} $$
Then we have, $$ a_{n} = \sum_{r=1}^{n}\frac{1}{r!} \ + 2\sum_{r=2}^{n}\frac{1}{r!} \ + 3\sum_{r=3}^{n}\frac{1}{r!} \ + \cdots  + n\sum_{r=n}^{n}\frac{1}{r!} \ \geq \ 1 + \sum_{r=1}^{n}\frac{1}{r!}$$ for all $n \geq 2$.
$\implies \displaystyle  \lim_{n \to \infty}a_{n} \ \geq \ e $
Now, in my book it says $\displaystyle  \lim_{n \to \infty}a_{n} \ = \frac{3}{2}e $
How can I attack this problem? Anyone please?
 A: Hint:
You know that $\displaystyle e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$ and that $\displaystyle \sum_{k=1}^n k = \frac{n(n+1)}{2}$.
So $\displaystyle \sum_{n=1}^{\infty}\frac{1+2+…+n}{n!} = \frac{1}{2}\sum_{n=0}^\infty \frac{n(n+1)}{n!}$.
Now what is the derivative of $xe^x$? Its second derivative? What is $e^1$?
A: $$\sum_{k=1}^n k = \frac{n(n+1)}{2}.$$  Thus the given sum is $$S = \frac{1}{2} \sum_{n=1}^\infty \frac{n(n+1)}{n!} = \frac{1}{2} \sum_{n=1}^\infty \frac{n+1}{(n-1)!} = \frac{1}{2} \left(\sum_{n=1}^\infty \frac{n-1}{(n-1)!} + 2\sum_{n=0}^\infty \frac{1}{n!}\right).$$  The second term in parentheses is simply $2e$; the first term simplifies further as $$\sum_{n=2}^\infty \frac{1}{(n-2)!} = e.$$  So the result is $$S = \frac{3e}{2}.$$
A: Note that
$$ \sum_{k = 1}^\infty \frac{1 + 2 + \cdots + k}{k!} = \frac{1}{2} \sum_{k = 1}^{\infty}  \frac{k + 1}{(k - 1)!} = \frac{1}{2} \left( \sum_{k = 1}^\infty \frac{(k - 1) + 1}{(k - 1)!} + e \right) = \frac{1}{2} \left( \sum_{k =2 }^\infty \frac{1}{(k - 2)!} + \sum_{k = 1}^{\infty} \frac{1}{(k-1)!} \right) + \frac{e}{2} = \frac{3e}{2} $$
A: Note that $1+2+\cdots+n=\frac{n(n+1)}{2}$.
Hence the prescribed sum is
$\frac{1}{2}\displaystyle\sum_{n=1}^{\infty}\frac{n^2+n}{n!}=\frac{1}{2}\displaystyle\sum_{n=1}^{\infty}\frac{n(n-1)+2n}{n!}=\frac{1}{2}\displaystyle\sum_{n=1}^{\infty}\frac{n(n-1)}{n!}+\displaystyle\sum_{n=1}^{\infty}\frac{n}{n!}.$
I hope you will be able to finish it from here. Good luck.
A: $$\sum_{i=1}^{n} i = \dfrac{n(n+1)}{2}$$ Hence we have:
$$\sum_{i=1}^{n} \dfrac{i(i+1)}{2\cdot i! } = \dfrac{1}{2}\sum_{i = 1}^{n} \dfrac{i+1}{(i-1)!} = \dfrac{1}{2} \sum_{i=0}^{n-1}\dfrac{i+2}{i!} \to \dfrac{3e}{2}$$
A: Since $e^x=\sum_{n\ge0}\frac{x^n}{n!}$,$$x^ke^x=x^k\frac{d^k}{dx^k}e^x=k!\sum_{n\ge0}\binom{n}{k}\frac{x^n}{n!},$$so$$\sum_{n\ge0}\underbrace{\frac{n(n+1)}{2}}_{\binom{n}{1}+\binom{n}{2}}\frac{x^n}{n!}=(x+x^2/2)e^x\implies\sum_{n\ge0}\frac{n(n+1)}{2}\frac{1}{n!}=3e/2.$$The sum starting at $1$ is the same, as the $n=0$ term is $0$.
A: First, let's note: $$(\forall n\in\mathbb{N}) \ \  \ 1+2+3+...+n= \frac{(n+1)n}{2} $$ so $$\eqalign{\sum_{1 \le n \le  \infty } \frac{1+2+...+n}{n!}&=\frac{1}{2} \sum_{1 \le n \le  \infty } \frac{n^2+n}{n!}=\cr &= \frac{1}{2} \sum_{1 \le n \le  \infty } \frac{n^2}{n!}+\frac{1}{2} \sum_{1 \le n \le  \infty } \frac{n}{n!}=\cr 
&=\frac{1}{2} \sum_{1 \le n \le  \infty }\frac{n}{(n-1)!}+\frac{1}{2} \sum_{1 \le n \le  \infty }\frac{1}{(n-1)!}=\cr &=\frac{1}{2} \sum_{0 \le k \le  \infty }\frac{k+1}{k!}+\frac{1}{2} \sum_{0 \le k \le  \infty }\frac{1}{k!}=\cr &=\frac{1}{2} \sum_{0 \le k \le  \infty }\frac{k}{k!}+\frac{1}{2} \sum_{0 \le k \le  \infty }\frac{1}{k!}+\frac{1}{2} \sum_{0 \le k \le  \infty }\frac{1}{k!}=\cr &= \frac{1}{2} \sum_{1 \le k \le  \infty }\frac{k}{k!}+\frac{1}{2} \sum_{0 \le k \le  \infty }\frac{1}{k!}+\frac{1}{2} \sum_{0 \le k \le  \infty }\frac{1}{k!}=\cr &= \frac{1}{2} \sum_{1 \le k \le  \infty }\frac{1}{(k-1)!}+\frac{1}{2} \sum_{0 \le k \le  \infty }\frac{1}{k!}+\frac{1}{2} \sum_{0 \le k \le  \infty }\frac{1}{k!}=\cr &=\frac{1}{2} \sum_{0 \le k \le  \infty }\frac{1}{k!}+\frac{1}{2} \sum_{0 \le k \le  \infty }\frac{1}{k!}+\frac{1}{2} \sum_{0 \le k \le  \infty }\frac{1}{k!}=\cr &=3\cdot\frac{e}{2} }$$
or
$$e^x= \sum_{n=0}^{ \infty }  \frac{x^n}{n!}$$ $$\frac{\text{d}}{\text{d}x}e^x= \sum_{n=0}^{ \infty }n\frac{x^{n-1}}{n!}$$ $$x^2\frac{\text{d}}{\text{d}x}e^x= \sum_{n=0}^{ \infty }n\frac{x^{n+1}}{n!}$$
$$\frac{\text{d}}{\text{d}x}\left(x^2\frac{\text{d}}{\text{d}x}e^x\right)= \sum_{n=0}^{ \infty }n(n+1)\frac{x^{n}}{n!}$$
$$\frac{\frac{\text{d}}{\text{d}x}\left(x^2\frac{\text{d}}{\text{d}x}e^x\right)}{2}= \sum_{n=0}^{ \infty }\frac{\frac{n(n+1)}{2}}{n!}x^{n}$$
count the left side at $x=1$
