Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

We know Ramanujan got this result $$\sqrt{1+2\sqrt{1+3\sqrt{1+\cdots }}}=3$$ and he used the formula $$x+n+a=\sqrt{ax+{{(n+a)}^{2}}+x\sqrt{a(x+n)+{{(n+a)}^{2}}+(x+n)\sqrt{\cdots }}}$$ where $x=2,n=1,a=0$ ,we get the first result, but I don't know how to prove it, can you help me?

share|improve this question

1 Answer 1

up vote 28 down vote accepted

$$(x+n+a)^2 = x^2 + n^2 + a^2 + 2an + 2ax + 2nx$$ $$ = ax + (n+a)^2 + x(x + a + 2n)$$

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

which you can substitute for $(x+n) + n + a$ again and again to get the sequence of iterated roots.

The convergence is because the sequence is monotone increasing but bounded above by $x+n+a$ for $n > 0$, $a,x \ge 0$ (after enough ($k$) iterations, $x + k*n + a$ will be greater than 1.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.