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Show that the series obtained from a generalization of the decimal expansion converges

by RoRi

We may generalize the decimal expansion of a number by replacing the integer 10 with any integer b> 1. If x > 0, let a_0 denote the greatest integer greater than x. Assuming the integers a_0, a_1, \ldots, a_{n-1} have been defined, let a, denote the largest integer such that

    \[ \sum_{k=0}^n \frac{a_k}{b^k} \leq x. \]

Show that the series

    \[ \sum_{k=0}^{\infty} \frac{a_k}{b^k} \]

converges and has sum x.

Proof. Since a_k \leq b-1 we have

    \[ \sum_{k=0}^{\infty} \frac{a_k}{b^k} \leq \sum_{k=0}^{\infty} \frac{b-1}{b^k} < \sum_{k=0}^{\infty} \frac{b}{b^k} = b \sum_{k=0}^{\infty} \frac{1}{b^k} = b \left( \frac{1}{1 - \frac{1}{b}} \right). \]

Since

    \[ 0 \leq \sum_{k=0}^{\infty} \frac{a_k}{b^k} < \frac{b^2}{b-1} \]

we have established the convergence of

    \[ \sum_{k=0}^{\infty} \frac{a_k}{b^k}. \qquad \blacksquare \]

Describe a geometric connection with a generalization of the decimal expansion of a number

by RoRi

We may generalize the decimal expansion of a number by replacing the integer 10 with any integer b> 1. If x > 0, let a_0 denote the greatest integer greater than x. Assuming the integers a_0, a_1, \ldots, a_{n-1} have been defined, let a, denote the largest integer such that

    \[ \sum_{k=0}^n \frac{a_k}{b^k} \leq x. \]

Describe a geometric method for obtaining a_0, a_1, a_2, \ldots.

Instead of dividing the real line into segments with 10 subintervals and taking the greatest integer number of intervals, we divide the line into b subintervals.

Prove properties of a generalization of the decimal expansion of a number

by RoRi

We may generalize the decimal expansion of a number by replacing the integer 10 with any integer b> 1. If x > 0, let a_0 denote the greatest integer greater than x. Assuming the integers a_0, a_1, \ldots, a_{n-1} have been defined, let a, denote the largest integer such that

    \[ \sum_{k=0}^n \frac{a_k}{b^k} \leq x. \]

Show that 0 \leq a_k \leq b-1 for each k \geq 1.

Proof. Suppose otherwise, that for some n \geq 1 we have a_n \geq b. Hence,

    \begin{align*} \sum_{k=0}^n \frac{a_k}{b^k} \leq x && \implies && \sum_{k=0}^{n=1} \frac{a_k}{b^k} + \frac{a_n}{b^n} &\leq x \\[9pt] && \implies && \sum_{k=0}^{n-1} \frac{a_k}{b^k} + \frac{(b+r)}{b^n} &\leq x \\[9pt] && \implies && \sum_{k=0}^{n-1} \frac{(a_k + 1)}{b^k} + \frac{r}{b^n} &\leq x \\[9pt] && \implies && \sum_{k=0}^{n-1} \frac{(a_k+1)}{b^k} &\leq x. \end{align*}

This contradicts that a_{n-1} is the greatest integer such that

    \[ \sum_{k=0}^{n-1} \frac{a_k}{b^k} \leq x. \qquad \blacksquare \]

Prove that decimals ending in zeros can also be written as a decimal ending in repeated nines

by RoRi

If the decimal expansion of a number ends in zeros, prove that this number can also be written as a decimal which ends in nines if we decrease the last nonzero digit in the decimal expansion by one unit. Prove this statement using infinite series.

Proof. Let

    \[ x = a_0. a_1 \ldots a_k a_{k+1} \ldots \]

where a_i = 9 for all i \geq k +1. Then we have,

    \begin{align*} && 10^k x &= a_0 a_1 \ldots a_k.a_{k+1} \ldots \\ &&&= R + \sum_{n=1}^{\infty} \frac{9}{10^n} \end{align*}

where R = a_0 a_1 \ldots a_k is an integer. Then,

    \begin{align*} 10^k x &= R + 9 \left( \sum_{n=0}^{\infty} \frac{1}{10^n} - 1 \right) \\ &= R + 9 \left( \frac{1}{1-\frac{1}{10}} - 1 \right) \\ &= R + 1 \\ \implies x &= \frac{R+1}{10^k} = a_0. a_1 \ldots a_{k-1}(a_k+1). \qquad \blacksquare \end{align*}