Why does this sum $\sum\limits_{n=1}^{\infty}{\frac{2^{2^{n-1}}}{2^{2^n}-1}}$ converge to 1? A friend of mine gave me this quiz: Where does this sum converge to?
$$\sum\limits_{n=1}^{\infty}{\frac{2^{2^{n-1}}}{2^{2^n}-1}}$$
$a_1=\frac{2}{3} , a_2=\frac{4}{15}, a_3=\frac{16}{255},...$and $S_1=\frac{2}{3}, S_2=\frac{14}{15} , S_3=\frac{254}{255}, ...$
So I thought that this sum will converge to 1.

Let $$2^{2^{n-1}}=A_n, a_n=\frac{A_n}{A_n^2-1}=\frac{1}{2}\left(\frac{1}{A_n+1}+\frac{1}{A_n-1}\right)$$
  Since $A_n = A_{n-1}^2$, I thought that this could lead to finding the partial sum of the series. I could prove that $$S_n=1-\frac{1}{A_n^2-1}$$by mathematical induction, but I want to know whether my method can be useful for finding $S_n$ from scratch.

So my question is : Is there a way to find $S_n$ from scratch??
Also, I found that $$a_n=\frac{\frac{1}{A_n}}{1-\frac{1}{A_n^2}}=\sum_{n=1}^{\infty}{\frac{1}{A_n^{2n-1}}}=\frac{1}{2^{1\cdot2^{n-1}}}+\frac{1}{2^{3\cdot 2^{n-1}}}+\frac{1}{2^{5\cdot 2^{n-1}}}+....\\
 \therefore \sum_{n=1}^\infty {a_n}=\frac{1}{2^{1\cdot 2^{0}}}+\frac{1}{2^{3\cdot 2^{0}}}+\frac{1}{2^{5\cdot 2^{0}}}+...\\
+\frac{1}{2^{1\cdot 2^{1}}}+\frac{1}{2^{3\cdot 2^{1}}}+\frac{1}{2^{5\cdot 2^{1}}}+...\\ +\frac{1}{2^{1\cdot 2^{2}}}+\frac{1}{2^{3\cdot 2^{2}}}+\frac{1}{2^{5\cdot 2^{2}}}+...
\\ =\sum_{n=1}^{\infty}{2^{-n}}=1$$
Is this a valid way? I used the Riemann series theorem to rearrange them.

Here's another manipulation you could do that doesn't require any rearranging, just the simple fact that two convergent series can be added term-wise (which is a commutativity of addition plus an addition of convergent sequences statement). Let's write $$s_n = \frac{2^{2^{n-1}}}{2^{2^n}-1};$$
$$a_n = \frac{1}{2^{2^{n-1}}-1};$$
$$b_n = \frac{1}{2^{2^{n-1}}+1};$$
$S=\sum_{n=1}^\infty s_n$; $A = \sum_{n=1}^\infty a_n$; and $B = \sum_{n=1}^\infty b_n$. It's easy to see that $A,B < \infty$ by comparison to some simple (e.g., geometric) series. Since $s_n = a_n/2 + b_n/2$, we have that $S=A/2+B/2<\infty$ and the following hold:
$$a_n = a_{n-1}/2 - b_{n-1}/2, \;\;\;\mbox{so}$$
$$A = a_{0}/2-b_0/2 + A/2 - B/2 = 1+A/2-B/2, \;\;\;\mbox{so}$$
$$S = A/2 + B/2 = (1+A/2-B/2)/2+B/2 = 1/2+(A/2+B/2)/2 = 1/2 + S/2,$$
which of course implies $S=1$.
