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?

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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 $2+1=3$. :)

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 $k=1$.

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## Comments

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

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 $f(0) = 0 +1$.

I hope this is adequate.

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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 $2+1=3$. :)

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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 $k=1$.

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

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@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

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

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

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Thanks! :)

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

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$f(x)^{2}-1=xf(x+1)$

Yeah sure! I just checked tillLog in to reply

@Calvin Lin can clear it up)

Here is the issue (which I'm not sure if it's really an issue, maybeWhen going from $f(0)$ to $f(1)$ using the recurrence relation and solving for $f(1)$, what do we get?

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

Oh yeah!Log in to reply

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

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Is there any other possible approach?

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Not one that I know.

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Its value is 3

This is very famous nested radical given by ramanujan.

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

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