Ramanujan's nested radical is great!

Algebra Level 4

Let $S= \displaystyle\sum_{x=0}^{62}\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}}$

Hence, find $\lfloor S\rfloor$

Hints: $3=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}$

$2+1=\sqrt{1+2\sqrt{1+(2+1)\sqrt{1+(2+2)\sqrt{1+\cdots}}}}$

1 solution

Tommy Li
May 26, 2016

$x+1=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+...}}}}$ (by Ramanujan)

$\sum_{i=0}^{62}\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+...}}}}=1+2+3+...+63=\frac{(63+1)(63)}{2}=2016$

$\lfloor 2016\rfloor =2016$