Showing the series of real decimal is convergent If $a_n\ldots a_0.a_{-1} a_{-2}\ldots$ is a real decimal. How do I show that the series $\sum_ia_i10^i$ is absolutely convergent, where the sum is over natural numbers $i$ such that $-\infty < i \leq n$.
I am struggling with starting the proof by finding the partial sums to show the sequence is convergent. I am having trouble with the fact that the index goes to $-\infty$, so could anyone please help me with this?
Appreciate any help.
 A: Because the infinite bound is a negative infinity you're going to take partial sums starting at $n$ and working your way down instead of up.  So your first few partial sums are given by the following decimal numbers:
$$s_n = a_n0\ldots0$$
$$s_{n - 1} = a_na_{n-1}0\ldots0$$
$$\vdots$$
$$s_0 = a_n\ldots a_0$$
$$s_{-1} = a_n\ldots a_0.a_{-1}$$
$$s_{-2} = a_n\ldots a_0.a_{-1}a_{-2}$$
Everything in sight is positive so once we prove this converges then it automatically converges absolutely.  To see that it converges I would suggest using a monotone convergence theorem.  You know that the $s_n$'s are increasing monotonically so you just need to show that they're bounded above, for example you could show $s_i \leq (a_n + 1)10^n$ for all $i$.
A: For $K>0$ let $Y_K=a_n...a_0. a_{-1},...a_{-K}.$ For $p,q \in \Bbb N$ we have $$0\leq Y_{K+p+q}-Y_{K+p}=\sum_{j=1}^qa_{K+p+j}10^{-(K+p+j)}\leq$$ $$\leq \sum_{j=1}^q9\cdot 10^{-(K+p+j)}=9\cdot 10^{-(K+p)}\sum_{j=1}^q10^{-j}=$$  $$=9\cdot10^{-(K+p)} \cdot 10^{-1}\cdot \frac {1-10^{-q}}{1-10^{-1}}=$$ $$=10^{-(K+p)}(1-10^{-q})<10^{-(K+p)}.$$ 
Given $\epsilon >0$, take $K\in \Bbb N$ large enough that $10^{-(K+1)}<\epsilon.$ If $n_1$ and $n_2$ are greater than $K$, with $n_1\ne n_2$, then $\{n_1,n_2\}=\{K+p, K+p+q\}$ for some $p,q\in \Bbb N.$ 
So for all $n_1,n_2$ greater than $K$ we have $|Y_{n_1}-Y_{n_2}|<10^{-(K+p)}\leq 10^{-(K+1)}<\epsilon.$
