I have the following result, but without a proof:

$$ g_n = \frac{1}{1-2^{-n}} = (1+2^{-n})(1+2^{-2n})(1+2^{-4n})\cdots = \prod_{j=0}^{\infty}(1+2^{-2^jn}), $$

To prove that the result is correct I defined the sequence

$$ a_k(n) = \prod_{j=0}^k(1+2^{-2^jn}) = a_{k-1}(n)(1+2^{-2^kn}) \Rightarrow \\a_k(n) = a_{k-1}(n)(1+2^{-2^kn}) $$

So my attempt was trying to show that $g_n$ the solution of the sequence above, when $k$ diverges.

This is equivalent to show that $a_k(n) \sim g_n$. But I got stuck and I don't know what to do now... I'm also quite sure that there's an easier solution than the one I'm proposing.


I had another clue... given that

$$ g_n = \sum_{j=0}^{+\infty} (2^{-n})^j = lim_{k\rightarrow +\infty} \sum_{j=0}^{2^k-1} (2^{-n})^j = lim_{k\rightarrow +\infty} S_k $$

I observe that

$$ S_k = \sum_{j=0}^{2^k-1} (2^{-n})^j = \sum_{j=0}^{2^{k-1}-1} (2^{-n})^j + \sum_{j=2^{k-1}}^{2^k-1} (2^{-n})^j = \\ \sum_{j=0}^{2^{k-1}-1} (2^{-n})^j + 2^{-2^{k-1}n} \sum_{j=0}^{2^{k-1}-1} (2^{-n})^j = (1+ 2^{-2^{k-1}n})\sum_{j=0}^{2^{k-1}-1} (2^{-n})^j = (1+ 2^{-2^{k-1}n})S_{k-1} $$

So I end up with a sequence:

$S_k = S_{k-1}(1+ 2^{-2^{k-1}n}) \;,\; S_k \rightarrow \frac{1}{1-2^{-n}}$,

Is this correct? (Sorry for my notation)



\begin{align}(1 - 2^{-n})(1 + 2^{-n}) &= 1-2^{-2n}, \\ (1 - 2^{-2n})(1 + 2^{-2n}) &= 1 - 2^{-4n},\end{align} and so on.


$$(1 - 2^{-n})(1 + 2^{-n})(1 + 2^{-2n})\cdots(1 + 2^{-2^jn}) = 1 - 2^{-2^{j+1}n},$$


$$(1 + 2^{-n})(1 + 2^{-2n})\cdots (1 + 2^{-2^jn}) = \frac{1-2^{-2^{j+1}n}}{1 - 2^{-n}}$$

Taking the limit as $j\to \infty$, we obtain the result.


But if you multiply $g_n = (1+2^{-n})(1+2^{-2n})(1+2^{-4n})\cdots $ one gets $$g_n=(1+2^{-n}+2^{-2n}+2^{-3n}+2^{-4n}+2^{-5n}+2^{-6n}+2^{-7n})(1+2^{-8n})....$$ which in turns form a infinite GP with first term $a=1$ and common ratio $r=2^{-n}$ $$\frac{a}{1-r}$$


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