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sqrt(1 + sqrt(2 + sqrt(3 + sqrt(4 + ...)))) = 1.75793275661800...

"It was discovered by T. Vijayaraghavan that the infinite radical, sqrt( a_1 + sqrt( a_2 + sqrt ( a_3 + sqrt( a_4 + ... )))) where a_n >= 0, will converge to a limit if and only if the limit of log(a_n)/2^n exists." [Clawson, 229; cf. A072449].

We know the asymptotic limit of primes and hence that the Prime Nested Radical converges.

Clawson misstates Vijayaraghavan's theorem. Vijayaraghavan proved that for a_n > 0, the infinite radical sqrt(a_1 + sqrt(a_2 + sqrt(a_3 + ...))) converges if and only if limsup (log a_n)/2^n < infinity. (For example, suppose a_n = 1 if n is odd, and a_n = e^2^n if n is even. Then (log a_n)/2^n = 0, 1, 0, 1, 0, 1, ... for n >= 1, so the limit does not exist. However, limsup (log a_n)/2^n = 1 and the infinite radical converges.) - Jonathan Sondow, Mar 25 2014

The continued fraction expression of this is A105818. "It was discovered by T. Vijayaraghavan that the infinite radical, sqrt( a_1 + sqrt( a_2 + sqrt ( a_3 + sqrt( a_4 + ... where a_n => 0, will converge to a limit if and only if the limit of (ln a_n)/2^n exists." [Clawson, 229; Sloane]. We know the asymptotic limit of Fibonacci numbers is Phi^n (Binet expansion) and that Phi^n < 2^n and hence that the Fibonacci Nested Radical converges.

Clawson misstates Vijayaraghavan's theorem. Vijayaraghavan proved that for a_n > 0, the infinite radical sqrt(a_1 + sqrt(a_2 + sqrt(a_3 + ...))) converges if and only if limsup (log a_n)/2^n < infinity. (For example, suppose a_n = 1 if n is odd, and a_n = e^2^n if n is even. Then (log a_n)/2^n = 0, 1, 0, 1, 0, 1, ... for n >= 1, so the limit does not exist. However, limsup (log a_n)/2^n = 1 and the infinite radical converges.) - Jonathan Sondow, Mar 25 2014

The semiprime nested radical is defined by the infinite recursion: sqrt(4 + sqrt(6 + sqrt(9 + sqrt(10 + sqrt(14 + ... + sqrt(semiprime(n))))). This converges by the criterion of T. Vijayaraghavan that "the infinite radical, sqrt( a_1 + sqrt( a_2 + sqrt ( a_3 + sqrt( a_4 + ... where a_n => 0, will converge to a limit if and only if the limit of (ln a_n)/2^n exists." [Clawson, 229; Sloane A072449]. The continued fraction representation of this constant is A105816.

Clawson misstates Vijayaraghavan's theorem. Vijayaraghavan proved that for a_n > 0, the infinite radical sqrt(a_1 + sqrt(a_2 + sqrt(a_3 + ...))) converges if and only if limsup (log a_n)/2^n < infinity. (For example, suppose a_n = 1 if n is odd, and a_n = e^2^n if n is even. Then (log a_n)/2^n = 0, 1, 0, 1, 0, 1, ... for n >= 1, so the limit does not exist. However, limsup (log a_n)/2^n = 1 and the infinite radical converges.) - Jonathan Sondow, Mar 25 2014

The semiprime nested radical is defined by the infinite recursion: sqrt(4 + sqrt(6 + sqrt(9 + sqrt(10 + sqrt(14 + ... + sqrt(semiprime(n))))). This converges by the criterion of T. Vijayaraghavan that "the infinite radical, sqrt( a_1 + sqrt( a_2 + sqrt ( a_3 + sqrt( a_4 + ... where a_n => 0, will converge to a limit if and only if the limit of (ln a_n)/2^n exists." [Clawson, 229; Sloane A072449].

Clawson misstates Vijayaraghavan's theorem. Vijayaraghavan proved that for a_n > 0, the infinite radical sqrt(a_1 + sqrt(a_2 + sqrt(a_3 + ...))) converges if and only if limsup (log a_n)/2^n < infinity. (For example, suppose a_n = 1 if n is odd, and a_n = e^2^n if n is even. Then (log a_n)/2^n = 0, 1, 0, 1, 0, 1, ... for n >= 1, so the limit does not exist. However, limsup (log a_n)/2^n = 1 and the infinite radical converges.) - Jonathan Sondow, Mar 25 2014

The decimal expansion of this is A105817. "It was discovered by T. Vijayaraghavan that the infinite radical, sqrt( a_1 + sqrt( a_2 + sqrt ( a_3 + sqrt( a_4 + ... where a_n => 0, will converge to a limit if and only if the limit of (ln a_n)/2^n exists." [Clawson, 229; Sloane]. We know the asymptotic limit of Fibonacci numbers is Phi^n (Binet expansion) and that Phi^n < 2^n and hence that the Fibonacci Nested Radical converges.

Clawson misstates Vijayaraghavan's theorem. Vijayaraghavan proved that for a_n > 0, the infinite radical sqrt(a_1 + sqrt(a_2 + sqrt(a_3 + ...))) converges if and only if limsup (log a_n)/2^n < infinity. (For example, suppose a_n = 1 if n is odd, and a_n = e^2^n if n is even. Then (log a_n)/2^n = 0, 1, 0, 1, 0, 1, ... for n >= 1, so the limit does not exist. However, limsup (log a_n)/2^n = 1 and the infinite radical converges.) - Jonathan Sondow, Mar 25 2014