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As in the title: $$ \lim_{n\to\infty} n \left(e-\sum_{k=0}^{n-1} \frac{1}{k!}\right) = ? $$ Numerically, it seems 0, but how to prove/disprove it? I tried to show that the speed of convergence of the sum to e is faster than $1/n$ but with no success.

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    $\begingroup$ You can try to show that $\frac{1}{n!}\leq\frac{2}{2^n}$, then $\frac{1}{n!}+\frac{1}{(n+1)!}+...\leq\frac{4}{2^n}$. $\endgroup$
    – Element118
    Nov 30, 2015 at 13:57
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    $\begingroup$ You can get more with $$\frac{1}{n!} < e - \sum_{k = 0}^{n-1} \frac{1}{k!} = \sum_{k = n}^{\infty} \frac{1}{k!} < \frac{1}{n!}\sum_{m = 0}^{\infty} \frac{1}{(n+1)^m} = \frac{1}{n!}\cdot \frac{n+1}{n},$$ namely $n!\bigl(e - \sum_{k = 0}^{n-1}\frac{1}{k!}\bigr) \to 1$. $\endgroup$ Nov 30, 2015 at 15:01

6 Answers 6

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The binomial coefficient

$${n+m\choose n}={(n+m)!\over n!m!}$$

is a positive integer if $m,n\ge0$, which implies

$${1\over(n+m)!}\le{1\over n!m!}$$

so

$$e-\sum_{k=0}^{n-1}{1\over k!}=\sum_{k=n}^\infty{1\over k!}=\sum_{m=0}^\infty{1\over(n+m)!}\le{1\over n!}\sum_{m=0}^\infty{1\over m!}={e\over n!}$$

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You can try to show that $\frac{1}{n!}\leq\frac{2}{2^n}$, then $\frac{1}{n!}+\frac{1}{(n+1)!}+\dots\leq\frac{4}{2^n}$.

$$\frac{1}{n!}\leq\frac{1}{1\times2\times3\times\dots\times n}\leq\frac{1}{1\times2\times2\times\dots\times2}=\frac{2}{2^n}$$

So $\frac{1}{n!}+\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\dots\leq\frac{2}{2^n}+\frac{2}{2^{n+1}}+\frac{2}{2^{n+2}}+\dots\leq\frac{4}{2^n}$.

Hence,

$$\lim_{n\rightarrow\infty}n(e-\sum_{k=0}^{n-1}\frac{1}{k!})\leq\lim_{n\rightarrow\infty}\frac{4n}{2^n}=0$$

Also, it is clear that the limit is non-negative, as $e-\sum_{k=0}^{n-1}\frac{1}{k!}\geq0$.

Hence, the limit of the expression as $n\rightarrow\infty$ is $0$.

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  • $\begingroup$ perfect argument. (+1) $\endgroup$
    – user98186
    Jan 27, 2016 at 15:51
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Late answer because something similar was tagged as "duplicate" (although it is not that what a "duplicate" is).

So,I adjust the answer from there to the question here:

Taylor gives

  • $e^1 = \sum_{k=0}^{n-1}\frac{1}{k!} + \frac{e^{\xi_n}}{n!}$ with $\xi_n \in (0,1)$

It follows

$$0\leq n(e-\sum_{k=0}^{n-1} \frac{1}{k!}) = n\frac{e^{\xi_n}}{n!} \leq \frac{e}{(n-1)!} \stackrel{n \to \infty}{\longrightarrow}0$$

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HINT: (if you're familiar with the gamma function):

$$\lim_{n\to\infty}n\left(e-\sum_{k=0}^{n-1}\frac{1}{k!}\right)=$$ $$\lim_{n\to\infty}n\left(e-\frac{e\Gamma(n,1)}{\Gamma(n)}\right)=$$ $$\lim_{n\to\infty}n\left(\frac{e\left(\Gamma(n)-\Gamma(n,1)\right)}{\Gamma(n)}\right)=$$ $$\lim_{n\to\infty}\frac{en\left(\Gamma(n)-\Gamma(n,1)\right)}{\Gamma(n)}=$$ $$e\lim_{n\to\infty}\frac{n\Gamma(n)-n\Gamma(n,1)}{\Gamma(n)}=$$ $$e\lim_{n\to\infty}\left(n-\frac{n\Gamma(n,1)}{\Gamma(n)}\right)=$$ $$e\lim_{n\to\infty}\left(n-\frac{n^2\Gamma(n,1)}{n!}\right)=$$ $$-e\lim_{n\to\infty}\frac{n^2\Gamma(n,1)-nn!}{n!}$$

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I tried to show that the speed of convergence of the sum to e is faster than $1/n,$ but with no success.

Have you tried Stirling's approximation ?


In my opinion, a far more interesting question would have been trying to prove that

$$\lim_{n\to\infty}(n+a)\bigg[~e^b-\bigg(1+\dfrac b{n+c}\bigg)^{n+d}~\bigg]~=~b~e^b~\bigg(\dfrac b2+c-d\bigg),$$

since, in this case, the growth of the latter formula towards $e^b$ is comparable to that of $1/n.$
In your example, however, in order for the product to converge to a “meaningful” non-zero
quantity, the order of the multiplication factor should have been somewhere in the range of
$n!$, as has already been pointed put by Daniel Fischer in the comments.

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I will show that $\lim_{n \to \infty} n^m(e-\sum_{k=0}^{n-1} \frac{1}{k!}) = 0 $ for any finite $m$.

$\begin{array}\\ e-\sum_{k=0}^{n-1} \frac{1}{k!} &=\sum_{k=n}^{\infty} \frac{1}{k!}\\ &=\frac1{n!}\sum_{k=n}^{\infty} \frac{n!}{k!}\\ &=\frac1{n!}\sum_{k=n}^{\infty} \frac{1}{\prod_{j=n+1}^{k} j}\\ &=\frac1{n!}\sum_{k=n}^{\infty} \frac{1}{\prod_{j=1}^{k-n} (j+n)}\\ &\le\frac1{n!}\sum_{k=n}^{\infty} \frac{1}{\prod_{j=1}^{k-n} (1+n)}\\ &=\frac1{n!}\sum_{k=n}^{\infty} \frac{1}{(n+1)^{k-n}}\\ &=\frac1{n!}\sum_{k=0}^{\infty} \frac{1}{(n+1)^{k}}\\ &=\frac1{n!}\frac{1}{1-\frac1{n+1}}\\ &=\frac1{n!}\frac{n+1}{n}\\ \end{array} $

Since $\lim_{n \to \infty} \frac{n^m}{n!} = 0 $ for any finite $m$, $\lim_{n \to \infty} n^m(e-\sum_{k=0}^{n-1} \frac{1}{k!}) = 0 $ for any finite $m$.

Also note that $\lim_{n \to \infty} n!(e-\sum_{k=0}^{n-1} \frac{1}{k!}) = 1 $.

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