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The sum of the series $1+\frac{1+2}{1!}+\frac{1+2+3}{3!}+....$ equals?

The answer is $\frac{3e}{2}$.

But I dont Know How?

I have tried following:

$1+\frac{1+2}{1!}+\frac{1+2+3}{3!}+....$=$\sum\limits_{n=1}^{\infty} \frac{n(n+1)}{2n!} $

I know that sequence of partial sum $S_n$=$1+\frac{1+2}{1!}+\frac{1+2+3}{3!}+....+\frac{n(n+1)}{2n!}$ and $\sum\limits_{n=1}^{\infty} \frac{n(n+1)} {2n!}$ converges to same point.

Now how to find $S_n$ converges to which point?

kalpeshmpopat
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  • There was a similar post yesterday, I can't find it now but the idea is use the Taylor Exapnsion of $e^x$, differentiate it twice and you will see what happens. – cr001 Dec 15 '15 at 08:22
  • Please check the exact definition of your series. The intended pattern remains unclear. – Christian Blatter Dec 15 '15 at 14:28

2 Answers2

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Hint: $$\frac{n(n+1)}{n!}=\frac{n+1}{(n-1)!}=\frac{(n-1)+2}{(n-1)!}=\frac{(n-1)}{(n-1)!}+2\frac{1}{(n-1)!}=\frac{1}{(n-2)!}+2\frac{1}{(n-1)!}$$ the both last ones with $n\to\infty$ gives us $e$

Zelos Malum
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You are considering $$S=\sum_{i=1}^\infty a_n$$ where $$a_n=\frac{\sum_{i=1}^n i}{n!}=\frac 12\frac{n (n+1)}{n!}$$ Now consider $$T(x)=\sum_{i=1}^\infty a_n x^n=\frac 12\sum_{i=1}^\infty \frac{n (n+1)}{n!}x^n$$ and write $n(n+1)=n(n-1)+2n$. So $$2T(x)=\sum_{i=1}^\infty \frac{n (n-1)}{n!}x^n+2\sum_{i=1}^\infty \frac{n}{n!}x^n=x^2\sum_{i=1}^\infty \frac{n (n-1)}{n!}x^{n-2}+2x\sum_{i=1}^\infty \frac{n}{n!}x^{n-1}$$ Now, remembering that $e^x=\sum_{i=1}^\infty \frac{x^n }{n!}$, you can notice that the first summation is the second derivative of $e^x$ (that is to say $e^x$) and that the second summation is the first derivative of $e^x$ (that is to say $e^x$ again). Replacing, $$2T(x)=x^2e^x+2xe^x$$ Now, use $x=1$.

Claude Leibovici
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