<s>[INST] 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?
 [/INST] 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 $x=1$
</s>