# How can this be an integer??

One of my friends told me that $$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+...}}}}}$$ comes out to be an integer. Can anyone help me out with guessing that single digit integer?

Note by Pranjal Jain
4 years ago

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The method to solve this kind of nested radical is to define a function $f(x)=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}}$ Clearly $f(x)^2-1=xf(x+1)$ Now note that $f(0)=1$ We will use induction to show that $f(x)=x+1$ for all positive integers $$x$$ (which will in turn give us the value of $$f(2)$$) First note the base case $f(0)=1=0+1$ Now suppose $f(k)=k+1$ for some positive integer $$k$$. Then $f(k)^2-1=kf(k+1)\implies (k+1)^2-1=kf(k+1)\implies k^2+2k=kf(k+1)\implies f(k+1)=k+2$ and the induction is done.

We conclude that $f(x)=x+1$ for all positive integers $$x$$.

Now we compute the value of $$f(2)$$ simply as $$2+1=3$$. :)

- 4 years ago

Thanks dude! Wikipedia had a more general one which was more difficult to be understood! Keep it up!

- 4 years ago

Thanks! :)

There is still a slight issue with the induction though. (I think). See if you can find it.

- 4 years ago

Yeah sure! I just checked till $$f(x)^{2}-1=xf(x+1)$$

- 4 years ago

Here is the issue (which I'm not sure if it's really an issue, maybe @Calvin Lin can clear it up)

When going from $$f(0)$$ to $$f(1)$$ using the recurrence relation and solving for $$f(1)$$, what do we get?

- 4 years ago

Oh yeah! $$\frac{0}{0}$$ indeterminate form!!!

- 4 years ago

Not quite. In essence, you have not proven anything as yet. You need the fact that $$f(1) = 2$$ to conclude that $$f(k) = k+1$$.

The induction step is only valid if you can divide by $$k$$ (right at the end). Hence, this need not hold true for $$k = 0$$. As such, your base case needs to be $$k=1$$.

Staff - 4 years ago

Can you help me in proving this? I guess @Nathan Ramesh 's method is not bad and "something" can be done with induction.

- 4 years ago

@Calvin Lin does this prove the base case?

$2=\sqrt{1+1\cdot 3}$ $\sqrt{1+1\cdot 3}=\sqrt{1+1\sqrt{1+2\cdot 4}}$ $\sqrt{1+1\sqrt{1+2\cdot 4}}=\sqrt{1+1\sqrt{1+2\sqrt{1+3\cdot 5}}}$

We can continue this indefinitely

- 4 years ago

This way you can directly prove mine one to be 3!!!!

$$3=\sqrt{1+2×4}=\sqrt{1+2\sqrt{1+3×5}}$$ and so on as you did in your solution!!

- 4 years ago

Yes, of course.

- 4 years ago

It is 3! And it is reported in http://en.wikipedia.org/wiki/Nested_radical

- 4 years ago

Is there any other possible approach?

- 4 years ago

Not one that I know.

- 4 years ago

Its value is 3

This is very famous nested radical given by ramanujan.

- 4 years ago

yes i also approached same question on this site, the key is ramanujan

- 4 years ago

From: $f(x)^2 -1 = xf(x+1)$ $( f(x)+1 ) ( f(x)-1 ) = xf(x+1)$

Suppose $$f(x) = x+1$$, then: $(x+1+1)(x+1-1)=x(x+2)$ $x(x+2) \equiv x(x+2)$

It appears that $$f(x)=x+1$$ satisfies the first equation for all real values of $$x$$. The special case of $$x=0$$ is also shown to be $$f(0) = 0 +1$$.