Sum and Product of Infinite Radicals

Solution 1:

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

Where $C \approx 1.7579$ (OEIS A072449) is the nested radical constant. No closed form of this constant is known.

$$\sqrt{1\sqrt{2\sqrt{3\sqrt{4\cdots}}}}=\sigma$$

where $\sigma \approx 1.6616$ (OEIS A112302) is Somos's Quadratic Recurrence Constant.

$\sigma$ has an alternate form (I hesitate to call it a "closed form") of $$\sigma = \exp\left[-2^n \frac{\partial \operatorname{Li_n\left(\frac{1}{2}\right)}}{\partial n}\bigg|_{n=0}+\frac{1}{2} \frac{\partial \Phi\left(\frac{1}{2},s,n+1\right)}{\partial s}\bigg|_{s=0}\right]$$

where $\operatorname{Li}_n(z) = \sum^\infty_{n=1} \frac{z^k}{k^n}$ is the polylogarithm and $\Phi(z,s,a)=\sum_{k=0}^\infty \frac{z^k}{(a+k)^s}$ is the Lerch transcendent, and

$$\sigma = \exp\left[\int_0^1 \frac{1-x}{(x-2)\log x}\, \mathrm{d}x\right]$$

Solution 2:

For the first:

Well, if it converges, we may compute its logarithm. Thus, $\log x = \sum_{k=1}^\infty \frac{\log k}{2^k}$ Now, looking at this, it is evident that this actually DO converge, (the numerators are eventually dominated by $1.5^k$ so $\log x < K + \sum_{k=1}^\infty \frac{1.5^k}{2^k}$ for some finite constant $K.$ Now, this sum is geometric and converges).

Now, computing the VALUE of this is trickier, and it is most probably not a rational number or some "nice" expression. In fact, $x$ converges to about 1.66169 (Mathematica), and is the exponential of some expression involving a Polylogarithm.

EDIT: Mixing recursion and different arithmetic operations usually result in constants that have no nice expressions. For example, $1/F_1 + 1/F_2 + 1/F_4+\dots$ converges to a number which is unknown to be rational or not, irrational $F_i$ is the i'th Fibonacci number, http://en.wikipedia.org/wiki/Reciprocal_Fibonacci_constant