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Is there a closed form for this series:

$$\sum_{n=1}^\infty \left(e-\left(1+\frac{1}{n}\right)^n \right)^2 \approx 1.273278374727530507449$$

(Mathematica computation by Patrick Stevens).

This is basically a sum of squared errors for all $n$ for the classic limit used to define $e$.


Are there some other similar series of interest, representing a sum of squared errors? (I'm aware we can buid an infinite set of such series by using various limits for various constants, but I'm asking only about well known series).


I don't really have motivation except for the fact that this series seems fundamental enough to have been studied before.

Besides, there exists a special value for an infinite product:

$$\prod_{k=2}^{\infty} e \left(1-\frac{1}{k^2} \right)^{k^2}=\frac{\pi}{e^{3/2}}$$

(The link had been here, but it's broken now).


Update

Some attempts to rearrange the series:

$$e=\sum_{k=0}^\infty \frac{1}{k!}$$

$$\left(1+\frac{1}{n}\right)^n=\sum_{k=0}^n \left( \begin{array}( n \\ k \end{array} \right) \frac{1}{n^k}$$

Thus, we can write the general term as:

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

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  • $\begingroup$ It is at least convergent, by Raabe's test, but I have no idea how Mathematica calculated the relevant limit :P $\endgroup$ – Patrick Stevens Sep 14 '17 at 7:41
  • $\begingroup$ @PatrickStevens, thank you for the convergence confirmation, though it's WA, not Mathematica, I haven't been able to check with Mathematica because my license expired $\endgroup$ – Yuriy S Sep 14 '17 at 7:42
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    $\begingroup$ If I request 30 digits of precision, Mathematica gives me 1.273278374727530507449. $\endgroup$ – Patrick Stevens Sep 14 '17 at 7:42
  • $\begingroup$ @PatrickStevens, may I use this value in the OP? $\endgroup$ – Yuriy S Sep 14 '17 at 7:43
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    $\begingroup$ I took the liberty to add a link for the "special value of an infinite product". Hope it's OK. $\endgroup$ – Paramanand Singh Sep 14 '17 at 11:33
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This is not an answer.

I have been a bit surprised by the result given by Wolfram Alpha. The first result I got from it was $1.26411$ and asking for more digits $1.2686765$ which corresponds to what you wrote.

I computed the partial sums $$S_p=\sum_{n=1}^{10^p}\left(e-\left(1+\frac{1}{n}\right)^n \right)^2 $$ and got the following numbers $$\left( \begin{array}{cc} p & S_p \\ 1 & 1.111547861 \\ 2 & 1.255063881 \\ 3 & 1.271433724 \\ 4 & 1.273093674 \end{array} \right)$$

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  • $\begingroup$ Thank you! I edited the question using the value provided in the comments $\endgroup$ – Yuriy S Sep 14 '17 at 7:45
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Only some hints.

Be $\,W(x)\,$ the main branch of the Lambert W-function .

Based on my answer for Value of the series $\sum_\limits{n=1}^{\infty}\left(\frac{n+1}{n\cdot 2\pi}\right)^ n$ we get:

$\displaystyle f(x):=\sum\limits_{n=1}^\infty x^{n-1}\left(1+\frac{1}{n}\right)^n=-\int\limits_0^\infty \left(\frac{d}{dx}\left(\frac{W(-xte^{-t})}{xte^{-t}(1+W(-xte^{-t}))}\right)\right) dt$

$\displaystyle g(x):=\sum\limits_{n=1}^\infty x^{n-1}\left(1+\frac{2}{n}\right)^n=\int\limits_0^\infty \left(\frac{d}{dx}\left(\frac{W(-xte^{-t})^2}{(xte^{-t})^2(1+W(-xte^{-t}))}\right)\right) dt$

We have $\enspace \displaystyle \sum\limits_{n=1}^\infty x^{2n-2}\left(e-\left(1+\frac{1}{n}\right)^n\right)^2 = \frac{e^2}{1-x^2} – 2ef(x^2)+\frac{g(x)-g(-x)}{2x} $

and therefore $\enspace\displaystyle \sum\limits_{n=1}^\infty \left(e-\left(1+\frac{1}{n}\right)^n\right)^2 = \lim\limits_{x\uparrow 1} \left(\frac{e^2}{1-x^2} – 2ef(x^2)+\frac{g(x)-g(-x)}{2x}\right) \,$ .

Note: $\enspace$ Please understand that I don't have fun to calculate this. :-)

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