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Consider the following series,

$$1+\frac{1+2}{2!}+\frac{1+2+3}{3!}+\cdots$$

My Efforts

The series can be written in compact for as,

$$\sum_{n=1}^{\infty}{n(n+1)\over 2n!}$$

$$={1\over 2}(\sum_{n=1}^{\infty}{n^2\over n!}+\sum_{n=1}^{\infty}{n \over n!})$$ $$={1\over 2}(\sum_{n=1}^{\infty}{n^2\over n!}+\sum_{n=1}^{\infty}{1 \over (n-1)!})$$

Now $$\sum_{n=1}^{\infty}{1 \over (n-1)!}=1+1+{1\over 2!}+{1\over 3!}+\cdots=e$$ First summand can be simplified as, $$\sum_{n=1}^{\infty}{n \over (n-1)!}$$

But where does it converge? Any hints?

Edit: Thank you for all the quick replies. I now realize it was so easy. I just had to use $n=n-1+1$.

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    $\begingroup$ Your sum is equal to $$\frac{3}{2}e$$ $\endgroup$ – Dr. Sonnhard Graubner Sep 9 '18 at 9:45
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Hint:

$$ \sum_{n=1}^{\infty}{n \over (n-1)!} = \sum_{n=1}^{\infty}{n - 1 + 1 \over (n-1)!} $$

$$ = \sum_{n=2}^{\infty}{1 \over (n-2)!} + \sum_{n=1}^{\infty}{ 1 \over (n-1)!} $$

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\begin{align} \sum_{n=1}^{\infty}{n(n+1)\over 2n!} &= \sum_{n=1}^{\infty}{n-1+2\over 2(n-1)!} \\ &= \sum_{n=2}^{\infty}\dfrac{1}{2(n-2)!}+\sum_{n=1}^{\infty}\dfrac{1}{(n-1)!} \\ &= \dfrac{1}{2}e+e \\ &= \color{blue}{\dfrac{3}{2}e} \end{align}

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Hint: $\sum_{k=1} \frac{n^2}{n!} x^n =e^x (x^2 + x)$.

This can easily be proven by considering the Taylor expansion of $e^x$, multiplying, and making the change of variable.

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To me, the series is summarized as $$\sum_1^{\infty}\frac{n+1}{2(n-1)!}$$ instead! Now, use the ratio test.

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  • $\begingroup$ How does the ratio test tell where the series converges to? $\endgroup$ – GoodDeeds Sep 9 '18 at 9:48
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HInt: when $n\geq 2$, $$\frac{n } {(n-1)!}=\frac{n-1+1}{(n-1)!}=\frac{1}{(n-2)!}+\frac{1}{(n-1)!}$$

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$\sum_{n=1}^{\infty}\frac{n}{\left(n-1\right)!}=1+\sum_{n=2}^{\infty}\frac{\left(n-1\right)+1}{\left(n-1\right)!}=1+\sum_{n=0}^{\infty}\frac{1}{n!}+\sum_{n=1}^{\infty}\frac{1}{n!}=1+e+(e-1)=2e$

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