Calculus – Sequences and series – Remarkable decimal fractions

As an example of a geometric series we had already seen that \(0,\overline{1}=0,11111\ldots=\dfrac{1}{9}\). This can be done easier:

\[x=0,\overline{1}\quad\Longrightarrow\quad 10x=1,\overline{1}=1+0,\overline{1}=1+x\quad\Longrightarrow\quad 9x=1 \quad\Longrightarrow\quad x=\frac{1}{9}.\]

Stewart §11.2, Exercise 49
Consider the number \(x=0,\overline{9}=0,99999\ldots\), then we have \(10x=9,\overline{9}=9+0,\overline{9}=9+x\) and therefore: \(9x=9\) oftewel \(x=1\).

Of course this can also be done using a geometric series: \(x=\displaystyle\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\cdots=\sum_{n=1}^{\infty}\frac{9}{10^n}\). This is a geometric series with common ratio \(r=\displaystyle\frac{1}{10}\).
Since \(|r| < 1\) this series is convergent and its sum is \(\displaystyle\frac{\text{first term}}{1-\text{common ratio}}=\frac{\frac{9}{10}}{1-\frac{1}{10}}=1\).

Now we will prove that \(\displaystyle\frac{1}{81}=\frac{1}{9^2}=0,\overline{012345679}\).

We start with the geometric series \(\displaystyle\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}\) for \(|x| < 1\). Differentiation now leads to \(\displaystyle\sum_{n=1}^{\infty}nx^{n-1}=\frac{1}{(1-x)^2}\) for \(|x| < 1\). This implies that \(\displaystyle\sum_{n=1}^{\infty}nx^{n+1}=\frac{x^2}{(1-x)^2}=\left(\frac{x}{1-x}\right)^2\) for \(|x| < 1\). Substitution of \(x=\displaystyle\frac{1}{10}\) leads to \(\displaystyle\sum_{n=1}^{\infty}\frac{n}{10^{n+1}}=\left(\frac{1}{9}\right)^2=\frac{1}{81}\).

For the first nine terms we have

\[\sum_{n=1}^9\frac{n}{10^{n+1}}=0,01+0,002+0,0003+\cdots+0,0000000009=0,0123456789.\]

For the tenth term we have: \(\displaystyle\frac{10}{10^{11}}=\frac{1}{10^{10}}=0,0000000001\). Hence:

\[\sum_{n=1}^{10}\frac{n}{10^{n+1}}=0,0123456789+0,0000000001=0,0123456790.\]

Now suppose that \(p=0,\overline{012345679}\), then we have:

\[10^9p=12345679,\overline{012345679}=12345679+0,\overline{012345679}=12345679+p.\]

Since we have \(10^9-1=999999999=9^2\cdot12345679\), this implies that

\[(10^9-1)p=12345679\quad\Longleftrightarrow\quad9^2\cdot12345679p=12345679\quad\Longleftrightarrow\quad p=\frac{1}{9^2}=\frac{1}{81}.\]

Similarly, the substitution \(x=\displaystyle\frac{1}{100}\) leads to \(\displaystyle\sum_{n=1}^{\infty}\frac{n}{100^{n+1}}=\left(\frac{1}{99}\right)^2=\frac{1}{9801}\).

For the first ninety-nine terms we now have

\[\sum_{n=1}^{99}\frac{n}{100^{n+1}}=0,0001020304\ldots9596979899.\]

For the hundredth term we have: \(\displaystyle\frac{100}{100^{101}}=\frac{1}{100^{100}}\). Hence:

\[\sum_{n=1}^{100}\frac{n}{100^{n+1}}=0,0001020304\ldots9596979900.\]

Now suppose that \(q=0,\overline{0001020304\ldots95969799}\), then we have:

\[100^{99}q=1020304\ldots95969799,\overline{0001020304\ldots95969799}=1020304\ldots95969799+q.\]

Since we have \(100^{99}-1=99^2\cdot1020304\ldots95969799\), this implies that

\[(100^{99}-1)q=1020304\ldots95969799\quad\Longleftrightarrow\quad99^2\cdot1020304\ldots95969799q=1020304\ldots95969799 \quad\Longleftrightarrow\quad q=\frac{1}{99^2}=\frac{1}{9801}.\]

Similarly if follows that

\[\frac{1}{998001}=\frac{1}{999^2}=0,\overline{00001002003004\ldots995996997999},\] \[\frac{1}{99980001}=\frac{1}{9999^2}=0,\overline{000001000200030004\ldots9995999699979999},\] \[\frac{1}{9999800001}=\frac{1}{99999^2}=0,\overline{0000001000020000300004\ldots99995999969999799999}\]

and so on.

---

Last modified on March 15, 2021