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Calculate $$\sum_{r=1}^{20} (-1)^r\frac{r^2+r+1}{r!}\,.$$

I broke the sum into partial fractions and after writing 3-4 terms of the sequence I could see that it cancels but I wasn't able to arrive at the exact expression. I understand that it's trivial but for whatever reason I just couldn't get it so just need a little help.

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1 Answer 1

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$\mathcal{Hint}$

After writing $r=1$ separately, use the fact that the numerator can be written as $r(r-1)+2r+1$.

So, $$(-1)^r\frac{r^2+r+1}{r!}=(-1)^r\left(\frac{1}{(r-2)!}+\frac{2}{(r-1)!} +\frac{1}{r!}\right)$$

Can you finish now?

$\mathbf{Edit:-}$ Writing a few terms for $r=2$ onwards, we get

$$\left(\frac{1}{0!}+\frac{2}{1!}+\frac{1}{2!}\right)-\left(\frac{1}{1!}+\frac{2}{2!}+\frac{1}{3!}\right)+\left(\frac{1}{2!}+\frac{2}{3!}+\frac{1}{4!}\right)-...+\left(\frac{1}{16!}+\frac{2}{17!}+\frac{1}{18!}\right)-\left(\frac{1}{17!}+\frac{2}{18!}+\frac{1}{19!}\right)+\left(\frac{1}{18!}+\frac{2}{19!}+\frac{1}{20!}\right) $$

Notice how every term with denominator $2!,3!,...,18!$ completely cancel...Numerators of both $1$ and $19!$ are $2-1=1$ ...and we have the first and last terms $0!$ and $20$ as it is...

So your final answer is $\frac{1}{0!}+\frac{1}{1!}+\frac{1}{19!}+\frac{1}{20!}-3$.

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    $\begingroup$ The result of the summation up to $r=20$ is quite interesting. I hope they don't ask for the value ! Cheers $\endgroup$
    – Claude Leibovici
    Commented Aug 22, 2020 at 4:45
  • $\begingroup$ I already got this, I was asking for how to proceed after this. As I said in my post, I know it's trivial but the exact expression that telescopes is what eluded me $\endgroup$
    – Amadeus
    Commented Aug 22, 2020 at 5:21
  • $\begingroup$ +1 for the font in Hint. Can you suggest some books for evaluating series like this? $\endgroup$
    – Sujit Bhattacharyya
    Commented Aug 22, 2020 at 7:58
  • $\begingroup$ I don't know any books specific to this...This is just a trick I learnt from this site ..Writing $r^2=r(r-1)+r$ , $r^3=r(r-1)(r-2)+3r(r-1)+r$ also comes handy while solving summation of the type $r^2{ n\choose r}$, $r^3{n\choose r}$ etc. $\endgroup$
    – LM2357
    Commented Aug 22, 2020 at 8:00
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    $\begingroup$ @Sujit Bhattacharyya here you can just break it into 3 terms but more generally as the numerator has a power of 2 in it if we want to write it as a difference of terms then we would have to multiply two terms of the denominator. So let's say we multiply the last two terms in the denominator, $r$ and $r-1$. This gives $r²-r$. To get $r²+r+1$ we add $2r+1$. This was how I thought about it maybe someone else can add their logic as well $\endgroup$
    – Amadeus
    Commented Aug 22, 2020 at 9:56

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