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?

$$\sum_{k=1}^n k = \frac{n(n+1)}{2}.$$  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 $$S = \frac{3e}{2}.$$
