Show that $0.a_1a_2a_3\ldots = \sum_{n=1}^\infty \frac{a_n}{10^n}$ Show that $$0.a_1a_2a_3\ldots= \sum_{n=1}^\infty \frac{a_n}{10^n}$$
I always get confused by these simple questions.  Is it enough to just write out the terms of the sum and add them?  That seems way to easy for the real analysis course this is part of
 A: This is definition of the decimal expamsion of a real number and it cannot be proved, the right hand side tells us what the left hand side really means. 
We could be more precise and write it in this way to denote the base-dependence: $(0.a_1a_2a_3\ldots)_{10}= \sum_{n=1}^\infty \dfrac{a_n}{10^n}$.
I would surely not say that this below is a proof:
$(0.a_1a_2a_3\ldots)_{10}- \sum_{n=1}^\infty \dfrac{a_n}{10^n}=\sum_{n=1}^\infty \dfrac{a_n}{10^n} - \sum_{n=1}^\infty \dfrac{a_n}{10^n}=\sum_{n=1}^\infty ( \dfrac{a_n}{10^n}-\dfrac{a_n}{10^n})=\sum_{n=1}^\infty 0=0$
After first $=$ we use the definition of the decimal expansion of a real number, after second $=$ we use that the difference of convergent series is convergent and that that difference can be written in this way and have the same result if written in this way and after fourth $=$ we use the fact that the sum of an infinite number of zeros is zero.
But maybe your instructor had this style of reasoning if he wanted that you prove that, but I doubt it somehow.
A: Not sure if this what you meant, but we say $\sum a_n$ converges to $S$ if for every $\epsilon>0$ exists $N$ such that for every $n>N$ we have $\left|S-\sum a_n\right|<\epsilon$
Let $\epsilon$ be an arbitrary real number, then take $N=f(\epsilon)+1$ where $f(\epsilon)$ is the zero counting function (aka, it returns the number of zeros after the decimal point before meeting the first non-zero digit)
By definition of $f(\epsilon)$, we have $\left|S-\sum a_n\right|<\epsilon$
A: Given a finite string $(a_1,a_2,,\ldots, a_n)$ of digits, by definition
$$0.a_1a_2\ldots a_n:=\sum_{k=1}^n a_k\,10^{-k}\ .$$
Given an infinite sequence $(a_k)_{k\geq1}$ of numbers $a_k\in[0..9]$ the intended meaning of the infinite decimal expansion $0.a_1a_2a_3\ldots$ then is
$$0.a_1a_2a_3\ldots:=\sum_{k=1}^\infty a_k\,10^{-k}\ .\tag{1}$$
This is a definition, and not a theorem. What  needs proof in this context is the fact that the RHS of $(1)$ is well defined,  i.e., that the series converges in ${\mathbb R}$. Since its partial sums are increasing it is sufficient to note that
$$\sum_{k=1}^na_k\,10^{-k}\leq 9\sum_{k=1}^n10^{-k}={9\over10}{1-10^{-n}\over1-10^{-1}}<1\qquad(n\geq1)\ .$$
