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?

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

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 \(2+1=3\). :) – Nathan Ramesh · 2 years, 9 months ago

Log in to reply

– Pranjal Jain · 2 years, 9 months ago

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

There is still a slight issue with the induction though. (I think). See if you can find it. – Nathan Ramesh · 2 years, 9 months ago

Log in to reply

– Pranjal Jain · 2 years, 9 months ago

Yeah sure! I just checked till \(f(x)^{2}-1=xf(x+1)\)Log 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? – Nathan Ramesh · 2 years, 9 months ago

Log in to reply

– Pranjal Jain · 2 years, 9 months ago

Oh yeah! \(\frac{0}{0}\) indeterminate form!!!Log in to reply

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 \(k=1 \). – Calvin Lin Staff · 2 years, 9 months ago

Log in to reply

@Nathan Ramesh 's method is not bad and "something" can be done with induction. – Pranjal Jain · 2 years, 9 months ago

Can you help me in proving this? I guessLog in to reply

@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 – Nathan Ramesh · 2 years, 9 months ago

Log in to reply

\(3=\sqrt{1+2×4}=\sqrt{1+2\sqrt{1+3×5}}\) and so on as you did in your solution!! – Pranjal Jain · 2 years, 9 months ago

Log in to reply

– Nathan Ramesh · 2 years, 9 months ago

Yes, of course.Log in to reply

It is 3! And it is reported in http://en.wikipedia.org/wiki/Nested_radical – Chew-Seong Cheong · 2 years, 9 months ago

Log in to reply

– Pranjal Jain · 2 years, 9 months ago

Is there any other possible approach?Log in to reply

– Chew-Seong Cheong · 2 years, 9 months ago

Not one that I know.Log in to reply

Its value is 3

This is very famous nested radical given by ramanujan. – Krishna Sharma · 2 years, 9 months ago

Log in to reply

– Mardokay Mosazghi · 2 years, 9 months ago

yes i also approached same question on this site, the key is ramanujanLog in to reply

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. – Chew-Seong Cheong · 2 years, 9 months ago

Log in to reply