<s>[INST] 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.
 [/INST] 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 $S=1$.
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