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$$S=\sum_{k=2}^{n}\frac{k^{2}-2}{k!}, n\geq 2$$

I got $S=\sum_{k=2}^{n}\frac{1}{(k-2)!}+\frac{1}{(k-1)!}-\frac{1}{k!}-\frac{1}{k!}$

I give k values but not all terms are vanishing.I remain with $\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{(n-2)!}$

The sum should be $2-e+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{(n-2)!}$

DaniVaja
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    Why you think the sum should be $2-e+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{(n-2)!}$? Note that $S$ is a rational number but $2-e+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{(n-2)!}$ is't. – Qurultay May 25 '19 at 18:09
  • Because when I gave k values, I remained with S=2-e+ that sum – DaniVaja May 25 '19 at 18:10
  • And I don't know how to calculate that sum, to obtain a rational number – DaniVaja May 25 '19 at 18:10
  • See https://math.stackexchange.com/questions/576976/evaluate-the-series-lim-limits-n-to-infty-sum-limits-i-1n-fracn22 – lab bhattacharjee May 25 '19 at 18:20
  • @Qurultray get rid of that line of thinking - it's totally wrong. For example we have $$\sum_{k=1}^{\infty}k^{-2}=\frac{\pi^2}{6}\not\in\Bbb Q$$ while $k\in\Bbb N \implies k^{-2} \in \Bbb Q$ – Rhys Hughes May 25 '19 at 18:43
  • @RhysHughes Yes but this has to do with limits and the fact that $\mathbb Q$ is not complete. In the question there is no limit and hence the numbers are actually rationals (and turns out they also converge to rational number $3$). – P. Quinton May 25 '19 at 19:27

2 Answers2

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For the sum until the index $n$, you can apply the same trick : \begin{align*} \sum_{k=2}^n \frac{k^2-2}{k!} &= \sum_{k=2}^n \left[\frac{k (k-1)}{k!} + \frac{k}{k!}-\frac{2}{k!} \right]\\ &= \color{blue}{\sum_{k=2}^n \frac{k (k-1)}{k!}} + \color{orange}{\sum_{k=2}^n\frac{k}{k!}}-\color{green}{\sum_{k=2}^n\frac{2}{k!}}\\ &= \color{blue}{\sum_{k=0}^{n-2} \frac{1}{k!}} + \color{orange}{\sum_{k=0}^{n-2}\frac{1}{k!} - 1 + \frac{1}{(n-1)!}}- \color{green}{2\sum_{k=0}^{n-2}\frac{1}{k!} + 2+2-\frac{2}{n!}-\frac{2}{(n-1)!}}\\ &= 3-\frac{2}{n!}-\frac{1}{(n-1)!} \end{align*}

When $n$ tends to $\infty$, this converges to $3$.

P. Quinton
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    I think the question of the OP is about the sum from $2$ to $n$, not $2$ to $\infty$. One gets the limit you obtained plus an error term – Ewan Delanoy May 25 '19 at 18:08
  • In my book, the final answer is $S=2$ – DaniVaja May 25 '19 at 18:08
  • I don't understand the blue part transition.From k(k-1)/k! how you get just 1/k! ? What's the difference if there is n or infinity ? – DaniVaja May 25 '19 at 18:18
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    He got $\sum_{k=2}^\infty \frac{1}{(k-2)!}$, and then changed the initial index from $2$ to $0$. – Ishan Deo May 25 '19 at 18:23
  • @DaniVaja, fro the blue part: $k(k-1)/k!=1/(k-2)!$ then, make $j=k-2$ note that the indices of the sum changed from $k=0 to \infty$ to $j=2 to \infty$ – Julian Mejia May 25 '19 at 18:25
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$$ \eqalign{ & \sum\limits_{k = 2}^n {\left( {{1 \over {\left( {k - 2} \right)!}} + {1 \over {\left( {k - 1} \right)!}} - {2 \over {k!}}} \right)} = \sum\limits_{k = 0}^{n - 2} {{1 \over {k!}} + \sum\limits_{k = 1}^{n - 1} {{1 \over {k!}}} - \sum\limits_{k = 2}^n {{2 \over {k!}}} } = \cr & = 1 + 1 + \sum\limits_{k = 2}^{n - 2} {{1 \over {k!}}} + 1 + \sum\limits_{k = 2}^{n - 2} {{1 \over {k!}}} + {1 \over {\left( {n - 1} \right)!}} - 2\left( {\sum\limits_{k = 2}^{n - 2} {{1 \over {k!}}} + {1 \over {\left( {n - 1} \right)!}} + {1 \over {n!}}} \right) = \cr & = 3 - {1 \over {\left( {n - 1} \right)!}} - {2 \over {n!}} = 3 - {{n + 2} \over {n!}} \cr} $$ and it checks with the original sum, which in fact is rational and cannot include $e$.
So something is wrong somewhere in your notes.

G Cab
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