Suppose, we take a number between 0 and 1 which is x.
How can I write a limit of sum of this series:
sqrt(x) + sqrt(sqrt(x)) .. and so on until a square root limits to 1.
Getting a square root from a number between 0 and 1 on a calculator repeatedly gives us 1. Hence, is my question.
Consider that for any convergent infinite sum $\sum_{k = 1}^\infty a_k = L,$ we must have that
$$\begin{align}\lim_{n \to \infty} a_n & = \lim_{n \to \infty} \left(\sum_{k = 1}^n a_k - \sum_{k = 1}^{n - 1} a_k\right)\\ & = \lim_{n \to \infty} \sum_{k = 1}^n a_k - \lim_{n \to \infty} \sum_{k = 1}^{n - 1}a_k \\ & = L - L = 0\end{align}$$
So essentially, for an infinite sum to converge, the terms of the sum must converge to zero. Conversely, if the terms do not converge to zero, the sum must diverge. In calculus textbooks this is often called the $n^{th}$ term test for divergence, but it also goes by other names, such as the limit test, or simply the divergence test.
So, looking at our sum in particular, I claim that we have $\sum_{k = 1}^\infty x^{2^{-k}},$ for some real number $x$ with $0 < x < 1.$ If the way I've written out the terms with $k$ isn't clear, the point is that $a_1 = x^{2^{-1}} = x^{1/2} = \sqrt{x},$ and $$a_{k+1} = x^{2^{-k-1}} = x^{2^{-k} \cdot 2^{-1}} = \left(x^{2^{-k}}\right)^{1/2} = \sqrt{x^{2^{-k}}} = \sqrt{a_k}$$
so that each term is the square root of the one before it.
So, as stated before, for this sum to converge we would require the limit of the terms to go to zero, but when we compute the limit:
$$\lim_{n \to \infty} x^{2^{-n}} = \lim_{n \to \infty} e^{2^{-n} \ln(x)} = e^{\ln(x) \cdot \lim_{n \to \infty} 2^{-n}} = e^0 = 1$$
we get $1,$ as you indicated, which is clearly nonzero, meaning that the sum is divergent.