Can we make it equal to x?

Solution 1:

Why care must be taken

Equating any such infinite expression to a real number must be done with a hint of caution, because the expression need not have a real value. For example, setting $x = 1-1+1-1+\cdots$ is not correct, since one sees that $1-x = 1+(-1+1-1+1) = x$, so $x = \frac 12$ which is absurd from an addition point of view : whenever you truncate the series, it always has value either $1$ or $0$, so where does $\frac 12$ comes from? With this logic, it is safe to say $1-1+1-\cdots$ does not evaluate to any finite real number.

However, once we can confirm that the result of such an expression is real and well defined, then we can play with them as we do with real numbers.

Ok, so what about this one?

To confirm that $\sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$ is a finite real number, we need the language of sequences.

I won't go very far in, but essentially, if we define a sequence of reals by $x_1 = \sqrt 6$ and $x_{n+1} = \sqrt{6+x_n}$, then $x_ 2 = \sqrt{6+\sqrt 6}$, $x_3 = \sqrt{6+\sqrt{6+\sqrt 6}}$, and eventually, $x_n$ resembles more and more the expression that we are given to evaluate.

EDITED : I have modified the steps required for showing that $x_n$ is a convergent sequence, to the real number $3$.

• It is easy to see that $x_n$ is bounded. It is clearly positive for all $n$, and can be shown to be bounded by $3$ above by induction.

• $a_n$ is an increasing sequence can also be shown easily. Any bounded increasing sequence is convergent.

• Once convergence is shown, we can then assume that $\lim x_n = L$ exists, and then use continuity to take limits on $x_{n+1} = \sqrt{6+x_n}$ to see that $L = \sqrt{6+L}$. But $L \geq 0$ must happen. Thus, $L=3$ is the limit, and hence the value of the expression.

Versatility of sequences

To add to this, sequences also offer versatility. A similar question may be asked : what is: $$ 6+\cfrac{6}{6+\cfrac{6}{6+\cfrac{6}{6+\cdots}}} $$

What we do here is the same : use the language of sequences, by defining $x_1 = 6$ and $x_{n+1} = 6 + \frac{6}{x_n}$. Once again, we can check convergence i.e. that this quantity is a finite real number(But on this occasion, the sequence rather oscillates around the point of convergence before converging). Next, we can use limits to deduce that if $L$ is the value then it satisfies $L = 6+ \frac 6L$, which gives one reasonable candidate, $3+\sqrt{15}$. So, this expression is actually equal to $3+\sqrt{15}$.

It's not easy all the time!

However, the approach using sequences doesn't always give immediate rewards. For example, you could ask for the following : $$ \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}} $$

which also looks like a nested radical.Can we find a sequence which, for large $n$, looks like this expression? Try to write one down, which you can work with.

Anyway, the answer to the above expression is $3$! To see this, we need to use "reverse nesting" : $$ 3 = \sqrt{9} = \sqrt{1+2\cdot 4} = \sqrt{1+2\sqrt{16}} = \sqrt{1+2\sqrt{1+15}} \\ = \sqrt{1+2\sqrt{1+3\sqrt{25}}} = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{36}}}} \\= \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{49}}}}} =\cdots $$

And just breathe in, breathe out. Ramanujan, class ten I believe.

EDIT

The nested radical method is wrong, from a rigorous point of view, for the reason pointed out in the comments. However, there is a rigorous proof here. The proof by Yiorgos Smyrlis is brilliant.

Note that the "nested radical" method can be used for the earlier problem as well, by $3 = \sqrt{6+3} = \sqrt{6+\sqrt{6+3}} = \cdots$, but this is unrigorous, only providing intuition. You can try to see if something intuitive can be derived for the continued fraction.

Solution 2:

Let $a_1= \sqrt{6}$ and $a_{n+1}=\sqrt{6+a_n}$ for $n \ge 1$.

It is easy to see, by induction, that $(a_n) $ is increasing and that $0 \le a_n \le 3$ for all $n$. Hence $(a_n)$ is convergent. If $x$ is the limit of $(a_n)$, then we have $x^2-x-6=0$, thus $x=3$ or $x=-2$. Since all $a_n \ge 0$, we have $x=3.$