# Proving that $\left(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!}\right)$ has a limit

$$x_n=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!}$$

How can we prove that the sequence $(x_n)$ has a limit? I have to use the fact that an increasing sequence has a limit iff it is bounded from above. No more "advanced" tools can be used. It's obvious that this sequence is increasing, but I am having trouble finding a bound.

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You need to use the following result "If $\{ a_n \}$ is a monotone sequence of real numbers (i.e., if $a_n ≤ a_{n+1}$ or $a_n ≥ a_{n+1}$ for every $n ≥ 1$), then this sequence has a finite limit if and only if the sequence is bounded". –  Mhenni Benghorbal Jan 8 '13 at 1:11

$x_n>1+1+\frac12=2.5$

Nowas David Mitra suggested, $$\frac1{3!}=\frac1{1\cdot2\cdot3}<\frac1{2\cdot2}=\frac 1{2^2},$$ $$\frac1{4!}=\frac1{1\cdot2\cdot3\cdot4}<\frac1{2\cdot2\cdot2}=\frac 1{2^3}$$ $$\frac1{5!}=\frac1{1\cdot2\cdot3\cdot4\cdot5}<\frac1{2\cdot2\cdot2\cdot2}=\frac 1{2^4}$$

and so on.

Hence $$x_n<1+1+\frac12+\frac1{2^2}+\frac1{2^3}+\cdots< 1+\sum_{0\le r<\infty}\frac1{2^r}=1+\frac1{1-\frac12}=3$$

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Alternately you can use

$$x_n=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!} \leq x_n=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+...+\frac{1}{(n-1)\cdot n}$$

and

$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+...+\frac{1}{(n-1)\cdot n}$$ is telescopic, since

$$\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1} \,.$$

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I know you may not be able to use this for the task at hand, but I couldn't resist pointing out is really good to know in the long run:

The infinite sum given by $x_n$ defines (is one representation of) the mathematical constant e. That is,

$$\lim_{n\to \infty} x_n\,=\,\lim_{n\to \infty} \frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!} + ... \,=\,\sum_{n=0}^\infty \frac{1}{n!} \,= \,e.$$

So, indeed, as demonstrated by lab bhattacharjee, $$2.5 \;\lt\; \lim x_n = e \;\lt \;3$$

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Careful, you say that $x_n$ is a finite sum and saying that it equals an infinite sum. Furthermore you ambiguously use $n$. These are just small mistakes with the notation. How about saying $\lim_{n\to \infty} x_n = \lim_{n\to \infty} \frac{1}{0!}+\ldots+\frac{1}{n!}=\sum_{k=0}^{\infty}\frac{1}{k!}=e$? –  Christian Ivicevic Jan 8 '13 at 1:19
@ChristianIvicevic fair enough. I was intending to address the OPs question about the limit of $x_n$, but should stick to being precise and explicit. –  amWhy Jan 8 '13 at 1:29
$x_n=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}<1+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{n-1}}<1+\sum_{k=0}^\infty\frac{1}{2^k}$