How to show $$1/e=1-1+\frac1{2!}-\frac1{3!}+\frac1{4!}+\dots$$
I have consider the product of the nth partial sums of the expansions for $e$ and $1/e$ but still can't prove there are equal to 1
with the definition $e=1+1+\frac1{2!}+\frac1{3!}+\dots$
Asked
Active
Viewed 159 times
1
-
2How do you define $e$? – Simply Beautiful Art Oct 01 '17 at 12:22
-
by using sum to infinity of 1+1+1/2!+1/3!+... – Yen Chuan Oi Oct 01 '17 at 12:28
-
then you should include that in your post. Leaving it out results in answers such as Janitha357's, which use information about $e^x$ you have not. – Simply Beautiful Art Oct 01 '17 at 12:30
1 Answers
4
I assume you know that
$$e = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \ldots = \sum_{k=0}^{\infty} \frac{1}{k!}.$$
Now let
$$f = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \ldots = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!}.$$
By the Mertens theorem and the binomial formula
$$\begin{align*} e \cdot f & = \left( \sum_{k=0}^{\infty} \frac{1}{k!} \right) \cdot \left( \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \right) = \sum_{n=0}^{\infty} \sum_{k=0}^n \frac{(-1)^{n-k}}{k! (n-k)!} \\[1ex] & = \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^n \binom{n}{k} \cdot 1^k \cdot (-1)^{n-k} = \sum_{n=0}^{\infty} \frac{(1-1)^n}{n!} = \frac{1}{0!} = 1 \end{align*}$$
so $f = \frac{1}{e}$. In the end we used the fact that
$$0^n = \begin{cases} 1 & \text{ if } n = 0 \\ 0 & \text{ if } n > 0 \end{cases}$$
Adayah
- 10,468
-
-
@YenChuanOi Without proving that $$\sum_{n=0}^{\infty} \frac{x^n}{n!} = \lim_{n \to \infty} \left( 1+\frac{x}{n} \right)^n$$ at least for $x = 1$ and $x=-1$, I don't know of any. – Adayah Oct 01 '17 at 13:03
-