$\sqrt{6 +\sqrt{6 +\sqrt{6 + \ldots}}}$. This is the famous question. I have to calculate it's value. I found somewhere to the solution to be putting this number equal to a variable $x$. That is, $\sqrt{6 +\sqrt{6 +\sqrt{6 + \ldots}}} = x$.

Then we square both the sides. $6 +{\sqrt{6 +\sqrt{6 + \ldots}}} = x^2$.

Then we replace the square root thing with $x$.

$6 + x = x^2$ and solve the equation to get the answer as 3.

But I have a doubt that what type of number is this? Is it a real number or not? And if it isn't, how can we perform mathematical operations on it?

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    $\begingroup$ It's the limit of a sequence of real numbers, so, if it exists, it's real. $\endgroup$ – Gerry Myerson Nov 14 '18 at 11:49
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    $\begingroup$ $$3=\sqrt{6+3}=\sqrt{6+\sqrt{6+3}}=\cdots$$ $\endgroup$ – lab bhattacharjee Nov 14 '18 at 11:55
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    $\begingroup$ Such expressions are called Nested Radicals $\endgroup$ – Yadati Kiran Nov 14 '18 at 11:56
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    $\begingroup$ You have proven that if the sequence converges, then it converges to 3. The next step would be to prove that it actually does converge. $\endgroup$ – user3482749 Nov 14 '18 at 11:56
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    $\begingroup$ @GerryMyerson True. However, it's worth noting that one can easily make a mistake from similar (but incorrect) reasoning. For example, a limit of a sequence of rational numbers is not necessarily rational (but it is real!). $\endgroup$ – Eff Nov 14 '18 at 11:56

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 $x = 1-1+1-1+...$ 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 $1-1+1-...$ 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 $\sqrt{6+\sqrt{6+\sqrt{6+...}}}$ 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: $$ 6+\frac{6}{6+\frac{6}{6+\frac{6}{6+...}}} $$

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 : $$ \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}} $$

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" : $$ 3 = \sqrt{9} = \sqrt{1+2\cdot 4} = \sqrt{1+2\sqrt{16}} = \sqrt{1+2\sqrt{1+15}} \\ = \sqrt{1+2\sqrt{1+3\sqrt{25}}} = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{36}}}} \\= \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{49}}}}} =... $$

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


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 $3 = \sqrt{6+3} = \sqrt{6+\sqrt{6+3}} = ...$, but this is unrigorous, only providing intuition. You can try to see if something intuitive can be derived for the continued fraction.

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    $\begingroup$ "Anyway, the answer to the above expression is 3!" Thought you meant 3 factorial at first :) $\endgroup$ – BioPhysicist Nov 14 '18 at 20:12
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    $\begingroup$ @AaronStevens $3! \neq 3$! $\endgroup$ – Todd Sewell Nov 14 '18 at 20:27
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    $\begingroup$ @ToddSewell Yes I know this. $\endgroup$ – BioPhysicist Nov 14 '18 at 20:29
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    $\begingroup$ The "reverse nesting" approach seems suspicious. Any number $x$ has a finite expansion $x = \sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + 4\sqrt{1 + 5\sqrt{C}}}}}$ for some constant $C$, which we can keep extending arbitrarily as long as we don't care what happens to $C$. In this way you can justify that the infinite radical is equal to anything. $\endgroup$ – Misha Lavrov Nov 14 '18 at 23:59
  • $\begingroup$ @MishaLavrov The "nested radical" is wrong, for the reasons you have mentioned. I should add that in the answer above. Indeed, there is a way of working with the above quantity with the help of sequences. I shall add a link to the answer above. $\endgroup$ – Teresa Lisbon Nov 15 '18 at 6:02

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 $x=3.$

  • $\begingroup$ Thanks for making sure this series is convergent. $\endgroup$ – Aayush Aggarwal Nov 14 '18 at 12:45

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