Home » Featured » Solving Ramanujan’s Puzzling Problem

# Solving Ramanujan’s Puzzling Problem

### Start here

Consider a sequence of functions as follows:-

$f_1 (x) = \sqrt {1+\sqrt {x} }$
$f_2 (x) = \sqrt{1+ \sqrt {1+2\sqrt {x} } }$

$f_3 (x) = \sqrt {1+ \sqrt {1+2 \sqrt {1+3 \sqrt {x} } } }$

……and so on to

$f_n (x) = \sqrt {1+\sqrt{1+2 \sqrt {1+3 \sqrt {\ldots \sqrt {1+n \sqrt {x} } } } } }$

Evaluate this function as n tends to infinity.

Or logically:

Find

$\displaystyle{\lim_{n \to \infty}} f_n (x)$ .

### Solution

Ramanujan discovered

$x+n+a= sqrt{ax + (n+a)^2 +xsqrt{a(x+n)+(n+a)^2 +(x+n)sqrt{...}}}$

which gives the special cases

$x+1 = sqrt{1+xsqrt{1 + (x+1)sqrt{1 + (x+2)sqrt{1 + (x+2)sqrt{...}}}}}$

for $x=2, n=1$ and $a=o$

$3 = sqrt{1+2sqrt{1+3 sqrt{1+ 4sqrt{1+cdots}}}}$

Proof of Ramanujan’s nested radicals equation$(x+n+a)^2 = (ax + (n+a)^2+ x(x+2n+a)$

$(x+n+a) =sqrt{ (ax + (n+a)^2+ x(x+2n+a)}$

$(x+2n+a) = ((x+n)+n+a)$

$(x+2n+a) = sqrt{(a(x+n) + (n+a)^2+ (x+n)((x+n)+2n+a)}$

$(x+2n+a) =sqrt{ (a(x+n) + (n+a)^2+ (x+n)((x+3n+a)}$

Then keep replacing the the last part term by writing $(alpha x+beta n+a)$ in the form of $(((x+(beta-1)n)+n+a)$ to give

$x+n+a= sqrt{ax + (n+a)^2 +xsqrt{a(x+n)+(n+a)^2 +(x+n)sqrt{...}}}$

Which is another form of our problem and is referred as Ramanujan’s Nested Radical. Comparing these two expressions & assuming

=$X$, we can write the problem as:

$\displaystyle {\lim_{n \to \infty}} f_n (x)$

= $\sqrt {1+X}$

= $\sqrt {1+3}$

=$\sqrt {4}$

=$2$

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Thank you for your precious comments and suggestions.

1. Carl Andreas says:

I’m not entirely sure, but using C++ (with n = 1,000,000) I numerically evaluated it to the function f(x) = 2. But as I said, not quite sure!

2. You’re right..! After using google, I got this Link , which was also saying the same. But I wasn’t satisfied.

3. animesh says:

Answer is 3 according to formulation 27 at http://mathworld.wolfram.com/NestedRadical.html .

4. Dave Radcliffe says:

There are two slightly different versions of this nested radical, so you need to be careful.

The version posed by Ramanujan was
sqrt(1 + 2*sqrt(1 + 3*sqrt(1 + 4*sqrt(1 + … = 3

Your version is almost the same:
sqrt(1 + 1*sqrt(1 + 2*sqrt(1 + 3*sqrt(1 + … = sqrt(1 + 3) = 2.

5. Tagus from Reddit says:

Hi there,

Just a small typo — I think you meant to write the limit as n tends to infinity. On all of the limits you wrote in that article, you unfortunately said that x goes to infinity.

x_x

• Gaurav says:

x_x Corrected Now. Thanks.

6. wow – wouldnt have a clue where to start!

7. Udit Agarwal says:

I think you might be one of the best bloggers in India today. We are having a TEDx conference, and it would be great to have you as a Speaker. I am sure you can come up with a very interesting talk. Let me know however I can contact you.

8. karthik says:

Ramanujan always the best

9. I really love the version that starts 3= because it smells like a magic number but really implicates the architecture of the number system we use.

10. narendra hegade says:

really nice.