Can I have help with this problem?
$$\sum_{r=1}^\infty \frac{(r+1)^2}{r!}=5e-1.$$
How to prove this series problem: $\sum_{r=1}^\infty \frac{(r+1)^2}{r!}=5e-1$?
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$\begingroup$
sequences-and-series
exponential-function
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1The upper limit was to $\infty$ right? I couldn't quite tell from the picture. – 2017-01-06
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2Hint: $$(r+1)^2=r(r-1)+3r+1$$ – 2017-01-06
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0This might help you with your question: [What's the value of $\sum\limits_{k=1}^{\infty}\frac{k^2}{k!}$?](http://math.stackexchange.com/q/44113) And you might also have a look at other posts [linked there](http://math.stackexchange.com/questions/linked/44113). – 2017-01-06
1 Answers
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$$\sum_{r=1}^{\infty}\frac{(r+1)^2}{r!}=\sum_{r=1}^{\infty}\frac{r^2}{r!}+2\sum_{r=1}^{\infty}\frac{r}{r!}+\sum_{r=1}^{\infty}\frac{1}{r!}=\sum_{r=1}^{\infty}\frac{r}{(r-1)!}+2\underbrace{\sum_{r=1}^{\infty}\frac{1}{(r-1)!}}_{e}+\underbrace{\sum_{r=1}^{\infty}\frac{1}{r!}}_{e-1}$$ thus $$\sum_{r=1}^{\infty}\frac{(r+1)^2}{r!}=\sum_{r=1}^{\infty}\frac{r-1+1}{(r-1)!}+3e-1=\underbrace{\sum_{r=2}^{\infty}\frac{1}{(r-2)!}}_{e}+\underbrace{\sum_{r=1}^{\infty}\frac{1}{(r-1)!}}_{e}+3e-1=5e-1$$