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We know that $$0.12345\cdots = \frac{1}{10}+\frac{2}{10^2}+\frac{3}{10^3}+\frac{4}{10^4}+\frac{5}{10^5}\cdots = \sum_{k=1}^{\infty}\frac{k}{10^k}$$

prove that $0.12345\cdots$ is irrational .

I though if let $0.1345\cdots$ be rational like proof of $\mathbb e$.

also I compute that series

NOTE :if $|x| < 1$ then $$\sum_{k=1}^{\infty}x^k=\frac{1}{1-x} \implies \sum_{k=1}^{\infty}kx^k= \frac{x}{(1-x)^2}$$

$$\sum_{k=1}^{\infty}\frac{k}{10^k}=\sum_{k=1}^{\infty}k\left(\frac{1}{10}\right)^k= \frac{\frac{1}{10}}{\left(1-\frac{1}{10}\right)^2}=\frac{10}{81}=0.1234567912346679\cdots$$ which is rational.

please give me hints. I don't want answer buddies.

thanks a lot.

Aditya Hase
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erfan soheil
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    The sum is rational, you have basically proved it. A sort of related number $0.1234567891011121314\cdots$ is irrational, indeed transcendental. The proof of irrationality is easy. – André Nicolas Dec 02 '14 at 07:20
  • if i understand correct, you say this number is like $0.1234567891011121314\cdots$ – erfan soheil Dec 02 '14 at 07:24

1 Answers1

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I don't know how to give you a hint

You're talking about two different things

$0.12345678910111213\cdots$ known as $C_{10}$ (Champernowne constant) is a real transcendental constant

On the other hand You're talking about $$\begin{align} \frac{1}{10}+\frac{2}{10^2}+\frac{3}{10^3}+\frac{4}{10^4}+\frac{5}{10^5}\cdots &=\sum_{k=1}^{\infty}\frac{k}{10^k}=\frac{10}{81}\end{align}=0.\overline{123456790}$$ which is rational and

Clearly $\displaystyle \quad \quad \quad \quad \quad \quad 0.12345678910111213\cdots\neq0.\overline{123456790}$

Since you're interested in proof of irrationality A bit of googling yields Answer by André Nicolas

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Aditya Hase
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