# Series for $\frac 1e$

When I read book, I see this: $$\lim D_n=1-\frac{1}{1!}+\frac{1}{2!}-\cdots+\frac{(-1)^n}{n!}=\frac{1}{e}$$ But I don't know why we have? Can you explain?

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It should be $\dfrac{1}{e}$, standard power series. Also, usually $D_n$ is the number of derangements, and it is $\dfrac{D_n}{n!}$ which is given by your sum. –  André Nicolas Jan 7 '13 at 4:46
sorry :D I am wrong –  Haruboy15 Jan 7 '13 at 4:48
you need to add $\lim_{n\to \infty}$ after the first equal sign. –  Maesumi Jan 7 '13 at 4:56
what is the definition of $e$ given in the book? –  Mohan Jan 7 '13 at 4:56
Your identity should say $\lim D_n=\lim 1-\frac{1}{1!}+\frac{1}{2!}-\cdots+\frac{(-1)^n}{n!}=\frac{1}{e}$, with "$\lim$" appearing after the first "$=$". –  Michael Hardy Jan 7 '13 at 5:24

The Maclaurin series for $e^x$ is

$$e^x=\sum_{k\ge 0}\frac{x^k}{k!}\;;$$

it’s valid for all $x$. For $x=-1$ this becomes

$$e^{-1}=\sum_{k\ge 0}\frac{(-1)^k}{k!}=1-\frac1{1!}+\frac1{2!}-\frac1{3!}+-\ldots~.$$

Your numbers $D_n$ are just the partial sums of this infinite series.

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I think it is not a nice solution , when I come to the contest I can't use Maclaurin series –  Haruboy15 Jan 7 '13 at 4:51
@Haruboy15 Your claim follows immediately from the definition. This is the only solution. What more do you want? –  Potato Jan 7 '13 at 4:56
Oh it's terrible , it 's easy to understand that my book only says that "easy to see...." –  Haruboy15 Jan 7 '13 at 5:01
(+1): This answer is very simple! –  Parth Kohli Jan 7 '13 at 19:17
It is the definition of $exp(x)$ which is equal to $e^x$. –  Andre Jan 7 '13 at 19:17

As $$\lim_{n\to \infty}\left(1+\frac1n\right)^n=e$$ $$e^x=\lim_{n\to \infty}\left(1+\frac1n\right)^{nx}=\lim_{n\to \infty}\left(1+\frac{nx}{1!}\frac1n+\frac{nx(nx-1)}{2!}\frac1{n^2}+\frac{nx(nx-1)(nx-2)}{3!}\frac1{n^3}+\cdots\right)$$

The 1st term $t_0=1$

The $r$ th term (where $r\ge1$) $$t_r=\frac{nx(nx-1)\cdots(nx-r+1)}{r!n^r}=\frac1{r!}\prod_{0\le t\le r-1}(x-\frac tn)$$

$$\lim_{n\to \infty}t_r=\frac{x^r}{r!}$$

So, $$e^x=\lim_{n\to \infty}(1+\frac1n)^{nx}=\sum_{0\le r< \infty}\frac{x^r}{r!}$$

Alternatively, let $$y=\sum_{0\le r< \infty}\frac{x^r}{r!}$$

$$\frac{dy}{dx}=\sum_{0\le r< \infty}\frac{rx^{r-1}}{r!}=\sum_{1\le r< \infty}\frac{x^{r-1}}{(r-1)!}=y$$

So, $$\frac{dy}y=dx$$

Integrating we get, $$\ln y= x+c$$

or, $$y=e^{c+x}\implies e^{c+x}=y=\sum_{0\le r< \infty}\frac{x^r}{r!}$$

Putting, $x=0$ we get, $e^c=1\implies c=0\implies e^x=\sum_{0\le r< \infty}\frac{x^r}{r!}$

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Perhaps the book was talking about Derangements. In this answer it is shown that \begin{align} \frac{\mathcal{D}(n)}{n!} &=\sum_{k=0}^n(-1)^k\frac1{k!}\\ &\approx \frac1e \end{align} The series $$e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$$ shows that \begin{align} \left|\,e^{-1}-\frac{\mathcal{D}(n)}{n!}\right| &=\left|\sum_{k=n+1}^\infty(-1)^k\frac1{k!}\right|\\ &\le\frac1{(n+1)!} \end{align}

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