Question

The meaning of the decimal representation of a number\(0 \cdot {d_1}{d_2}{d_3}\)(where the digit\({d_1}\)is one of the numbers\(0,1,2,........9\)) is that

\(0 \cdot {d_1}{d_2}{d_3}{d_4}...... = \frac{{{d_1}}}{{10}} + \frac{{{d_2}}}{{{{10}^2}}} + \frac{{{d_3}}}{{{{10}^3}}} + \frac{{{d_4}}}{{{{10}^4}}} + ....\)show that this series always converges.

Short answer

To prove that\(\sum\limits_{n = 1}^\infty {\frac{{{d_n}}}{{{{10}^n}}}} \)is convergent.

Given\(0 \cdot {d_1}{d_2}{d_3}{d_4}...... = \frac{{{d_1}}}{{10}} + \frac{{{d_2}}}{{{{10}^2}}} + \frac{{{d_3}}}{{{{10}^3}}} + \frac{{{d_4}}}{{{{10}^4}}} + ....\)which is equal to\(\sum\limits_{n = 1}^\infty {\frac{{{d_n}}}{{{{10}^n}}}} \).

Step-by-step solution

Given data

We have,

\(\begin{aligned}0 \cdot {d_1}{d_2}{d_3}{d_4}...... &= \frac{{{d_1}}}{{10}} + \frac{{{d_2}}}{{{{10}^2}}} + \frac{{{d_3}}}{{{{10}^3}}} + \frac{{{d_4}}}{{{{10}^4}}} + ....\\ &= \sum\limits_{n = 1}^\infty {\frac{{{d_n}}}{{{{10}^n}}}} \end{aligned}\)

Since\({d_n} \le 9\).

\( \Rightarrow \frac{{{d_n}}}{{{{10}^n}}} \le \frac{9}{{{{10}^n}}}\)

Let\({a_n} = \frac{{{d_n}}}{{{{10}^n}}}\)and\({b_n} = \frac{9}{{{{10}^n}}}\).

\( \Rightarrow {a_n} = {b_n}\)

Prove that\(\sum\limits_{n = 1}^\infty  {\frac{{{d_n}}}{{{{10}^n}}}} \)is convergent:

Since\(\sum\limits_{n = 1}^\infty {{b_n}} \)is convergent by comparison test series, \(\sum\limits_{n = 1}^\infty {\frac{{{d_n}}}{{{{10}^n}}}} \)is also convergent.

Therefore\(\sum\limits_{n = 1}^\infty {\frac{{{d_n}}}{{{{10}^n}}}} \)is convergent.

Hence proved.