Real Analysis, Sequences in relation to Geometric Series and their sums

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SUMMARY

The discussion centers on demonstrating the existence and uniqueness of a sequence of integers $(a_n)$ that satisfies the equation $x = \sum\limits_{n=1}^{\infty} \frac{a_n}{p^n}$ for a natural number $p > 1$ and a real number $0 < x < 1$. It is established that this sequence is unique unless $x$ is of the form $\frac{q}{p^n}$, where exactly two sequences exist. Additionally, the convergence of the series to a real number $x$ is confirmed, with the sequence of partial sums $\left(\sum_{n=1}^{N} \frac{a_n}{p^n}\right)_{N=1}^\infty$ converging to $x$ as $N$ approaches infinity.

PREREQUISITES
  • Understanding of real analysis concepts, particularly sequences and series.
  • Familiarity with the properties of geometric series.
  • Knowledge of limits and convergence in mathematical analysis.
  • Basic understanding of integer sequences and their properties.
NEXT STEPS
  • Study the properties of geometric series and their convergence criteria.
  • Explore the concept of unique representations of real numbers in different bases, such as binary and decimal expansions.
  • Investigate the implications of sequences of integers in real analysis, focusing on their convergence behavior.
  • Learn about the method of proof by contradiction, particularly in the context of uniqueness proofs in mathematical analysis.
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Mathematics students, educators, and researchers interested in real analysis, particularly those focusing on sequences, series, and their convergence properties.

joypav
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I will state the problem below. I don't quite understand what I am needing to show. Could someone point me in the right direction? I would greatly appreciate it.

Problem:

Let p be a natural number greater than 1, and x a real number, 0<x<1. Show that there is a sequence $(a_n)$ of integers with $0 \leq a_n < p$ for each n such that
$ x = \sum\limits_{n=1}^{\infty} \frac{a_n}{p^n} $

and that this sequence is unique except when x is of the form $ \frac{q}{p^n} $, in which case there are exactly two such sequences.

Show that, conversely, if $(a_n)$ is any sequence of integers with $ 0 \leq a_n < p$, the series

$ \sum\limits_{n=1}^{\infty} \frac{a_n}{p^n} $

converges to a real number x with $ 0 \leq x \leq 1 $. If p=10, this sequence is called the decimal expansion of x. For p=2 it is called the binary expansion of x, and for p=3, the ternary expansion.
 
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For a start, let $a_1$ be the greatest integer such that $\dfrac{a_1}{p} \le x$. Having chosen $a_1,\ldots, a_{N-1}$, let $a_N$ be the greatest integer such that $\displaystyle \sum_{n = 1}^N \frac{a_n}{p^n} \le x$. Show that the sequence of partial sums $\displaystyle \left(\sum_{n = 1}^N \frac{a_n}{p^n}\right)_{N=1}^\infty$ converges to $x$.
 
Euge said:
For a start, let $a_1$ be the greatest integer such that $\dfrac{a_1}{p} \le x$. Having chosen $a_1,\ldots, a_{N-1}$, let $a_N$ be the greatest integer such that $\displaystyle \sum_{n = 1}^N \frac{a_n}{p^n} \le x$. Show that the sequence of partial sums $\displaystyle \left(\sum_{n = 1}^N \frac{a_n}{p^n}\right)_{N=1}^\infty$ converges to $x$.

Then we have,

$ 0 < x - \sum\limits_{n=1}^{N} \frac{a_n}{p^n} < \frac{1}{p^N} $

Take the $ lim_{n\rightarrow{\infty}} $,
$ \vert x - \sum\limits_{n=1}^{\infty} \frac{a_n}{p^n} \vert = 0 $

$ \implies \sum\limits_{n=1}^{\infty} \frac{a_n}{p^n} $ converges to x.

Is this correct? Now to show uniqueness do I proceed by way of contradiction?
 
Also, I don't understand the second part of the problem either. Now I am to find the x it converges to?
 

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