Ramanujan Sum

Number Theory Level 2

$\large \sqrt{1+2 \sqrt{1+3 \sqrt{1+4\sqrt{1+5 \dots}}}} = ?$

\infty

3 25 None of the others 4 6

2 solutions

Chew-Seong Cheong
Feb 27, 2019

Let $f(x) = x+1$ , then

$\begin{aligned} f(x) & = \sqrt{(x+1)^2} = \sqrt{x^2+2x+1} = \sqrt{1 + x((x+1)+1)} \\ & = \sqrt{1 + x \color{#3D99F6}f(x+1)} \\ & = \sqrt{1 + x\color{#3D99F6}\sqrt{1 + (x+1)f(x+2)}} \\ & = \sqrt{1 + x\sqrt{1 + (x+1)\sqrt{1+(x+2)f(x+3)}}} \\ & = \sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)\sqrt{1+(x+4)\dots}}}}} \\ \implies f(2) & = \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6\dots}}}}} \\ & = 2+1 = \boxed 3 \end{aligned}$

@chakravarthy b , don't use all cap "RAMANUJAN SUM" as title or in text. All cap in text is equivalent to shouting in voice which is rude. Include everything in a formula "=?" in LaTex. Use three dots ... or \dots or \cdots not two dots for continue to infinity. It is a standard so need not explain. The multipication sighs $\times$ are unnecessary.

Chew-Seong Cheong - 2 years, 3 months ago

@Chew-Seong Cheong ok

chakravarthy b - 2 years, 3 months ago

@Chew-Seong Cheong @Nnsv Abhiram

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

Using this equation, the answer to the question , obtained by setting x = 2, n = 1, and a = 0.

to get info press here