# How is $\sum_{n=0}^N\frac{a_n}{10^n}$ a Cauchy sequence?

I've been studying the construction of real numbers this week and i've read about cantor's construction using the Cauchy sequence and Dedekind's construction. Now the book i'm reading (classic set theory for guided independent study) gives a new kind of construction by decimal expansion. First it says:

"We are quite accustomed to writing numbers by their decimal expansions. An expansion of this sort is really an infinite series of the form $$\sum_{n=0}^\infty\frac{a_n}{10^n}$$"

then it says

"The definition of an infinite series says that this is the limit of the sequence of its partial sums $$\langle s_N\rangle$$, where $$\langle s_N\rangle =\sum_{n=0}^N\frac{a_n}{10^n}$$ where all $$a_n$$ are integers and $$a_n\in\{0,1,2,3,4,5,6,7,8,9\}$$"

now the part i don't understand:

"$$\langle s_N\rangle$$ is a Cauchy sequence of rationals, which connects decimal expansions to Cantor reals - each equivalence class in Cantor's definition contains such a sequence $$\langle s_N\rangle$$"

How is $$\sum_{n=0}^N\frac{a_n}{10^n}$$ a cauchy sequence? And how is this a sequence at all if it's a series? i thought series and sequences were two different things..

for example $$\sum_{n=0}^3\frac{n}{10^n}=0+0.1+0.02+0.03$$ how is this a cauchy sequence? Maybe it's because i've never studied real analysis before and that's why i'm struggling with this, can you guys help me out please?

• In a decimal expansion $0\leq a_n \leq 9$. If $a_n$'s are arbitrary integers your sequence need not be Cauchy. Jul 1 '20 at 9:17
• @KaviRamaMurthy I suspect the intention was to say $a_n\in \{0,1,2,3,4,5,6,7,8,9\}$ but it did not come out like that Jul 1 '20 at 9:19
• @Henry But OP is clearly taking about unbounded sequences in the example $a_n=n$ that was considered. Jul 1 '20 at 9:21

The generic term of the sequence is $$s_N=\sum_{n=0}^N\frac{a_n}{10^n}$$ with $$N\geq 0$$. In order to show that $$(s_N)_N$$ is a Cauchy sequence, note that for $$M\geq N\geq 0$$, $$0\leq s_M-s_N=\sum_{n=N+1}^M\frac{a_n}{10^n}\leq 9\sum_{n=N+1}^M\frac{1}{10^n}<9\sum_{n=N+1}^\infty\frac{1}{10^n}=\frac{1}{10^N}$$ where we used that fact that $$0\leq a_n\leq 9$$. Can you take it from here?
• By the way I am not sure if it is maybe more instructive to finish off directly with $$9\sum_{n=N+1}^M \frac1{10^n}=10^{-N}-10^{-M}$$ Jul 1 '20 at 9:31
$$s_0 = a_0$$$$s_1 = a_0 + a_1/10$$$$s_2 = a_0 + a_1/10 + a_2/100$$ $$\vdots$$ $$s_N = a_0 + a_1/10 + a_2/100 + \cdots + a_N/10^N.$$
Each of those numbers $$s_N$$ is a series (a series is a sum of numbers). The ordered collection of numbers $$s_0, s_1, s_2, s_3$$ etc. is a sequence. The limit $$\lim\limits_{N \to \infty} s_N$$ of this series exists, and the expression $$\sum\limits_{n=0}^{\infty} a_n/10^n$$ is commonly used to denote this limit. When denoted this way, the limit is referred to as an infinite series.