Ramanujanish

Number Theory Level 2

Compute:

1 + 2 1 + 3 1 + 4 1 + ⋯ \large \sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + 4\sqrt{1 + \cdots}}}}


The answer is 3.

Mihir Mallick by Mihir Mallick , 18, India

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2 solutions

According Ramanujan's formula,

x + n + a = a x + ( n + a ) 2 + x a ( x + n ) + ( n + a ) 2 + ( x + n ) a ( x + 2 n ) + ( n + a ) 2 + ( x + 2 n ) ⋯ x+n+a = \sqrt{ax+(n+a)^2+x\sqrt{a(x+n)+(n+a)^2+(x+n)\sqrt{a(x+2n)+(n+a)^2+(x+2n)\sqrt \cdots}}}

Putting x = 2 x=2 , n = 1 n=1 and a = 0 a=0 , we have:

2 + 1 + 0 = 0 + ( 1 + 0 ) 2 + 2 0 ( 1 + 1 ) + ( 1 + 0 ) 2 + ( 2 + 1 ) 0 ( 1 + 2 ) + ( 1 + 0 ) 2 + ( 2 + 2 ) ⋯ 3 = 1 + 2 1 + 3 1 + 4 1 + ⋯ \begin{aligned} 2+1+0 & = \sqrt{0+(1+0)^2+2\sqrt{0(1+1)+(1+0)^2+(2+1)\sqrt{0(1+2)+(1+0)^2+(2+2)\sqrt \cdots}}} \\ 3 & = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}} \end{aligned}

6 Helpful 0 Interesting 0 Brilliant 0 Confused
Mihir Mallick
Dec 26, 2017

Keep taking the partial sums, it converges to 3

0 Helpful 0 Interesting 0 Brilliant 0 Confused

There was some typing mistake in the previous problem. I sincerely apologise to everyone. And I thank everyone those who reported against the previous problem.

Mihir Mallick

Mihir Mallick - 3 years, 5 months ago

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