1

I think this will be an easy problem for you, but I do not see the solution.

I know that

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

Knowing this, how can I demonstrate this

$$\displaystyle\sum_{k=1}^{\infty} (-1)^{k-1}\frac{1}{k!}=1-\frac{1}{e}$$

this was my attempt

i) $$\displaystyle\sum_{k=0}^{\infty} \frac{1}{k!}=e$$ ii)$$\displaystyle\sum_{k=1}^{\infty} \frac{1}{k!}=1+e$$ iii)$$\Bigg[\displaystyle\sum_{k=1}^{\infty}\frac{1}{k!}\Bigg]^{-1}=\frac{1}{1+e}$$

And I'm stuck here. I certainly took the wrong path. Can you help me?

Muradin Bronzebeard
  • 339

2 Answers2

0

Hint. Starting with

$$\sum_{k=1}^{\infty} (-1)^{k-1}\frac{1}{k!}=1-\frac{1}{e}$$

Move the $1$ to the left:

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

Get rid of the minus sign:

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

Now, multiply the series for $e$ and $\frac{1}{e}$, the result should end up being $1$:

$$e\cdot\frac{1}{e} = \left(\sum_{k=0}^{\infty}\frac{1}{k!}\right)\cdot\left(\sum_{k=0}^{\infty} (-1)^k\frac{1}{k!}\right)\overset{?}=\dots\overset{?}=1$$

lisyarus
  • 15,517
0

Note that

$$\sum_{k=1}^{\infty} (-1)^{k-1}\frac{1}{k!}= \sum_{k=1}^{\infty} \frac1{(-1)} \frac{(-1)^{k} }{k!}= -\sum_{k=1}^{\infty} \frac{(-1)^{k} }{k!}=1-\sum_{k=0}^{\infty} \frac{(-1)^{k} }{k!}=1-\frac1e$$

Indeed

$$e^x=\sum_{k=0}^{\infty} {x^k\over k!} \implies \sum_{k=0}^{\infty} \frac{(-1)^{k} }{k!}=e^{-1}=\frac1e$$

user
  • 154,566