Finding the Sum of a series $\frac{1}{1!} + \frac{1+2}{2!} +\frac{1+2+3}{3!}+...$

Question

I need to find the sum of this series $$\dfrac{1}{1!} + \dfrac{1+2}{2!} + \dfrac{1+2+3}{3!}+\ldots$$ But somehow I am not even convinced this converges. I tried writing it as $\sum \dfrac{n(n+1)}{2(n!)}$. I also tried to rewrite it as a telescoping series. But I seem to be stuck. Some hints on where I am going wrong would help. I am supposed to get the answer as $\dfrac{3e}{2}$.

Answer 1

You may write

$$ \begin{align} \sum_1^{\infty} \frac{n(n+1)}{2(n!)}&=\sum_1^{\infty} \frac{n(n-1)+2n}{2(n!)}\\\\ &=\sum_2^{\infty} \frac{n(n-1)}{2(n!)}+\sum_1^{\infty} \frac{2n}{2(n!)}\\\\ &=\frac12\sum_2^{\infty} \frac{1}{(n-2)!}+\sum_1^{\infty} \frac{1}{(n-1)!}\\\\ &=\frac12\sum_0^{\infty} \frac{1}{n!}+\sum_0^{\infty} \frac{1}{n!}\\\\ &=\frac32e \end{align} $$

since $$ \sum_0^{\infty} \frac{x^n}{n!}=e^x, \quad x \in \mathbb{C}. $$

Answer 2

What I said

$$\sum \frac{n^2+n}{2 n!}=\frac{1}{2}\left(\sum \frac{n\color{red}{+1-1}}{(n-1)!} +\sum \frac{1}{(n-1)!}\right)=\frac{1}{2}\left(\sum\frac{1}{(n-1)!}+ \sum \frac{1}{(n-2)!} +\sum \frac{1}{(n-1)!}\right)$$

And if you know that $\sum \frac{1}{n!}=e$ then...

And you know too algebra of limits, i.e., you know that if $\lim_{x\to a} f(x)=L$ and $\lim_{x\to a} g(x)=R$ and $L,R \in \mathbb R$ then $\lim_{x\to a}(f(x)+g(x))=\lim_{x\to a} f(x) + \lim_{x\to a} g(x)$

I will add some more: you can rearrange terms because all these series have absolute convergence, if they converges, because they are only of positive terms: Cauchy criteria for absolute convergence says that $\sum |a_n| \to L, L\in\mathbb R$ then if the series $\sum a_n$ converges, then it converges absolutely and you can rearrange terms and they converges to the same value.