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?

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 \(k=1 \).

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 \(f(0) = 0 +1\).

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TopNewestThe 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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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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@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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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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@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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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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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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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 \(f(0) = 0 +1\).

I hope this is adequate.

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