<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] Note that
$$ \sum_{k = 1}^\infty \frac{1 + 2 + \cdots + k}{k!} = \frac{1}{2} \sum_{k = 1}^{\infty}  \frac{k + 1}{(k - 1)!} = \frac{1}{2} \left( \sum_{k = 1}^\infty \frac{(k - 1) + 1}{(k - 1)!} + e \right) = \frac{1}{2} \left( \sum_{k =2 }^\infty \frac{1}{(k - 2)!} + \sum_{k = 1}^{\infty} \frac{1}{(k-1)!} \right) + \frac{e}{2} = \frac{3e}{2} $$
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