TITLE: Prove that the infinite sum $\sum_{n=1}^{\infty} \frac{F_{n}}{ 10 ^ n }$ converges to a rational number
QUESTION [20 upvotes]: How do you prove that the following infinite sum
\begin{align}
     &0.1
\\+\;&0.01
\\+\;&0.002
\\+\;&0.0003
\\+\;&0.00005
\\+\;&0.000008
\\+\;&0.0000013
\\ \;&\quad\vdots
\end{align}
converges to a rational number?
Notice that the above sum can be written as 
$$\sum_{n=1}^{\infty} \frac{F_{n}}{ 10 ^  n }$$
where $F_{n} $ is a Fibonacci sequence.

REPLY [5 votes]: The method used by both MvG and Anastasiya-Romanova 秀 assumes without proof that the series converges. If the series does not converge, it still produces a value but that value has no worth.
Thus you first need to show that $\sum_{n=1}^{\infty} \frac{F_n}{10^n}$ converges at all. Of course you can just follow user296113's lead and use Binet's formula, reducing the series to a couple of obviously convergent geometric series.
But if you want to solve the problem using nothing more about the Fibonacci numbers than their basic definition, you can first prove that $F_n < 2^n$ for all $n$:
First note that $F_0 = 0 < 1 = 2^0$ and $F_1 = 1 < 2 = 2^1$.
Assume that $F_{n-1} < 2^{n-1}$ and $F_{n-2} < 2^{n-2}$. Then $$F_n = F_{n-1} + F_{n-2} < 2^{n-1} + 2^{n-2} = 3\cdot2^{n-2} < 4\cdot 2^{n-2} = 2^n$$
The result follows by induction.
From this we have that $$\sum_{n=1}^{N} \frac{F_n}{10^n} < \sum_{n=1}^{N} \frac{2^n}{10^n} = \sum_{n=1}^{N} \left(\frac{1}{5}\right)^n < \sum_{n=1}^{\infty} \left(\frac{1}{5}\right)^n$$
For all $N$. Since the RHS converges, your series forms an increasing sequence in $N$ that is bounded above, and thus also must converge.