Skip to main content
### Number and algebra

### Geometry and measure

### Probability and statistics

### Working mathematically

### For younger learners

### Advanced mathematics

# There's a Limit

Or search by topic

Age 14 to 18

Challenge Level

- Problem
- Student Solutions

Tony of State College Area High School, Pennsylvania, USA and Chi Kin of Saint Dominic's International School, Lisbon sent in excellent solutions and here is Chi Kin's:

What happens when successive terms are taken in the continued fraction

I calculated the first five terms

$2 + {3\over2} = {7\over2} = 3.5 $

$2 + \frac{3}{2+{3\over2}} = {20\over7} = 2.86 $

The third term is $3.05$

The fourth term is $2.95$

And the fifth term is $3.02$

Thus, when successive terms are taken, the results oscillate alternately above and below $3$.

Indeed, if the fraction goes on forever, we would get $3$.

Suppose the continued fraction is denoted as $F$. Thus

$$\eqalign{ F &=& 2 + \frac{3}{F} \\ F &=& \frac{2F + 3}{F} \\ F^2 - 2F - 3 &=& 0 \\ (F-3)(F+1) &=& 0.}$$Thus $F = 3$, because $F$ cannot be negative.