# Thread: Proof of irrationality of e using series

1. ## Proof of irrationality of e using series

1) If $e$ were rational, it would be of the form $e=\frac{p}{q}$, where $p$ and $q$ are positive integers. Select an integer $k$ such that $k\geq 3$ and $k\geq q$. Use Taylor's theorem to show that

$\frac{p}{q}=e=1+\frac{1}{1!}+\frac{1}{2!}+...+\fra c{1}{k!}+\frac{e^z}{(k+1)!}$

for some $z\in (0,1)$

Is this just the Taylor series about 0. and the end part is the Lagrange formula for the remainder?

2) Suppose that:

$s_k=1+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{k!}$

Show that $k!(e-s_k)$ is an integer

3) Show that $0

4) Conclude that $e$ is irrational

NOTE: I've posted my answer below. I just need someone to check whether I've done it correct. Thanks

2. Originally Posted by acevipa
1) If $e$ were rational, it would be of the form $e=\frac{p}{q}$, where $p$ and $q$ are positive integers. Select an integer $k$ such that $k\geq 3$ and $k\geq q$. Use Taylor's theorem to show that

$\frac{p}{q}=e=1+\frac{1}{1!}+\frac{1}{2!}+...+\fra c{1}{k!}+\frac{e^z}{(k+1)!}$

for some $z\in (0,1)$

Is this just the Taylor series about 0. and the end part is the Lagrange formula for the remainder?

Yes.

2) Suppose that:

$s_k=1+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{k!}$

Show that $k!(e-s_k)$ is an integer

Check separatedly $\,\,k!e\,,\,k!s_k$

3) Show that $0

For this I think you need to already know that $e<4$ and do some simple algebra.

4) Conclude that $e$ is irrational

Obvious from the above.

Tonio

.

3. Thanks... I think I kind of understand it now.

4. So can someone verify my working, to make sure I haven't missed anything.

2) $\frac{p}{q}=e=1+\frac{1}{1!}+\frac{1}{2!}+...+\fra c{1}{k!}+\frac{e^z}{(k+1)!}$ for some $z\in(0,1)$

$k!(e-s_k)=k!e-k!s_k$

$=k!\left(\frac{p}{q}\right)-k!\left(1+\frac{1}{1!}+\frac{1}{2!}+...+\frac{1}{k !}\right)$

$\Rightarrow\left(\frac{k!}{q}\right)p-\left(k!+\frac{k!}{1!}+\frac{k!}{2!}+...+\frac{k!} {k!}\right)$

Since $k\geq q\Rightarrow \frac{k!}{q}$ is an integer. Therefore since all the terms are integers, it follows that $k!(e-s_k)$ must also be an integer

3) $e-s_k=\frac{e^z}{(k+1)!}$ for some $z\in (0,1)$ from (1) (Taylor's theorem)

$\Rightarrow 0

$\Rightarrow 0<(e-s_k)<\frac{3}{(k+1)}$

Since $k\geq 3\Rightarrow \frac{3}{(k+1)}<1$

$\Rightarrow 0<(e-s_k)<1$

4) From (2), since $k!(e-s_k)$ is an integer, but since $0.

This gives us a contradiction, therefore, $e$ is irrational

,

,

### prove that e is irrational using Taylor's theorem

Click on a term to search for related topics.