5
$\begingroup$

I'm a first year undergrad maths student and I've come across this particular sequence, which I can't find a limit for.

$$\underbrace{\sqrt{10\sqrt{10\sqrt{10\sqrt{10\cdots}}}}}_{n \text{ times}}$$

I know this may be embarrassingly easy, but I can't think of a way to start. Can you share any hints?

$\endgroup$
3
  • $\begingroup$ Whatever the limit is, it must satisfy $x=(10x)^{1/2}$ (why?) Then simplify this. $\endgroup$
    – Ian
    Oct 3, 2015 at 22:28
  • $\begingroup$ Should you check that a limit exists first? $\endgroup$
    – fleablood
    Oct 3, 2015 at 22:42
  • $\begingroup$ One needs to assign meaning to the expression in the OP. As written, it is ambiguous. The rigorous way forward requires a recurrence relationship for the sequence and an initial value. The way that this is written, the initial value is concealed. Please see my answer and please let me know how I can improve it. I really want to give you the best answer I can. $\endgroup$
    – Mark Viola
    Oct 3, 2015 at 22:53

4 Answers 4

8
$\begingroup$

Well..:

$$\sqrt{10\sqrt{10\sqrt{10\sqrt{10\cdots}}}} = 10^{\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots}$$

$\endgroup$
3
  • $\begingroup$ Nope, nothing wrong with it. This is merely a sketch or hint, as the OP asked for $\endgroup$
    – Victor
    Oct 3, 2015 at 22:58
  • 1
    $\begingroup$ @Dr.MV This is rigorous (or can be made rigorous) if you use the fact that the function $f(x)=10^x$ is continuous and evaluate the infinite sum in the exponent as a limit of partial sums. $\endgroup$ Oct 3, 2015 at 23:32
  • $\begingroup$ @michaelburr The expression $10^{\sum_n 2^{-n}}$ is well-defined. The original expression is not. But if we use the former to define the latter, then I completely agree with you. $\endgroup$
    – Mark Viola
    Oct 4, 2015 at 2:21
2
$\begingroup$

We have the recurrence relationship

$$x_{n+1}=\sqrt{10x_n}\implies x_{n+1}^2=10x_n$$

If $x_0=0$, then $x_n=0$ for all $n$ and $\lim_{n\to \infty}x_n=0$.

If $x_0=10$, then $x_n=10$ for all $n$ and $\lim_{n\to \infty}x_n=10$.

If $10>x_0>0$, then clearly $0<x_n<x_{n+1}$ and $x_n$ is increasing. And since $x_0<10$, the sequence is clearly bounded above by $10$. Since $x_n$ is monotonically increasing and bounded above it converges. Let's call this limit $L$. Thus, $\lim_{n\to \infty}x_{n+1}=L=\lim_{n\to \infty}\sqrt{10x_n}=\sqrt{10L}\implies L=10$

If $10<x$, then clearly $x_n>x_{n+1}>0$ and $x_n$ is decreasing. And since $x_n>0$, the sequence is clearly bounded below by $0$. Since $x_n$ is monotonically decreasing and bounded below it converges. Let's call this limit $L$. Thus, $\lim_{n\to \infty}x_{n+1}=L=\lim_{n\to \infty}\sqrt{10x_n}=\sqrt{10L}\implies L=10$.

Putting it all together, we have

$$\lim_{n\to \infty}x_n= \begin{cases} 10&,x_0>0\\\\ 0&,x_0=0 \end{cases} $$

and we are done!

$\endgroup$
4
  • $\begingroup$ This should be the accepted answer, all the others only use intuitive nonrigurous approaches. $\endgroup$ Oct 3, 2015 at 22:57
  • $\begingroup$ That's pretty good. I had a TA once claim $ x =\sqrt{2}^{\sqrt{2}^{...}}= 2 $ because $\sqrt{2}^x = x$ would mean x = 2. This would put a stop to that right away. $\endgroup$
    – fleablood
    Oct 3, 2015 at 23:11
  • $\begingroup$ @yotengunlcd Thank you so much! That made my weekend. $\endgroup$
    – Mark Viola
    Oct 3, 2015 at 23:21
  • $\begingroup$ @fleablood Thank you. And yes, I agree with you. $\endgroup$
    – Mark Viola
    Oct 3, 2015 at 23:25
1
$\begingroup$

Hint: Let $x = \lim_{n \to \infty}\underbrace{\sqrt{10\sqrt{10\sqrt{10\sqrt{10}}}}...}_{n \text{ times}}\ $.

Now observe that $$\underbrace{\sqrt{10\sqrt{10\sqrt{10\sqrt{10}}}}...}_{n \text{ times}} = \sqrt{10\underbrace{\sqrt{10\sqrt{10\sqrt{10}}}...}_{n-1 \text{ times}}}$$

so that in the limit, you have $x = \sqrt{10 x}$.

$\endgroup$
0
1
$\begingroup$

Let $x=\sqrt{10\sqrt{10\sqrt{10\cdots}}}$. Then, $x^2=10\sqrt{10\sqrt{10\sqrt{10\cdots}}}=10x$. Therefore, $x^2=10x$, so $x=10$.

$\endgroup$
2
  • $\begingroup$ This is not quite the rigorous way to proceed. One needs to assign meaning to the expression as written. The rigorous way forward requires a recurrence that has an initial value. The way that this is written, the initial value is concealed. If it were $0$, then, clearly the answer would be $0$, not $10$ as this result produces. But, this is a typical incorrect approach that is seen often. $\endgroup$
    – Mark Viola
    Oct 3, 2015 at 22:51
  • $\begingroup$ Yes, it is not rigorous. One would need to let $a_0=1$ and $a_n=\sqrt{10a_{n-1}}$ and squeeze the result (as you do in your answer). $\endgroup$ Oct 3, 2015 at 23:21

Not the answer you're looking for? Browse other questions tagged or ask your own question.