Can I have help with this problem?
$$\sum_{r=1}^\infty \frac{(r+1)^2}{r!}=5e-1.$$
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1The upper limit was to $\infty$ right? I couldn't quite tell from the picture. – pancini Jan 06 '17 at 07:26
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2Hint: $$(r+1)^2=r(r-1)+3r+1$$ – Did Jan 06 '17 at 07:29
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This might help you with your question: What's the value of $\sum\limits_{k=1}^{\infty}\frac{k^2}{k!}$? And you might also have a look at other posts linked there. – Martin Sleziak Jan 06 '17 at 09:06
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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 $$\sum_{r=1}^{\infty}\frac{(r+1)^2}{r!}=\sum_{r=1}^{\infty}\frac{r-1+1}{(r-1)!}+3e-1=\underbrace{\sum_{r=2}^{\infty}\frac{1}{(r-2)!}}_{e}+\underbrace{\sum_{r=1}^{\infty}\frac{1}{(r-1)!}}_{e}+3e-1=5e-1$$
Behrouz Maleki
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