I have that $$2^{(a+1)!}(1+\frac{1}{2^{1}}+\frac{1}{2^{2}}+...)=2(2^{a!})^{a+1}\,\,\,\,\,\,\,\,(1)$$ Which I am trying to show is $\geq \sum_{b=a+1}^{\infty}2^{b!}$ in order to prove $\sum_{b=0}^{\infty}2^{b!}$ is transcendental by Liouville's theorem. However I am struggling to understand how we get equality in $(1)$.
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2**Hint :** $\left(1+\dfrac{1}{2^{1}}+\dfrac{1}{2^{2}}+...\right)= 2$ – The Demonix _ Hermit Dec 14 '19 at 16:53

1Have you read the descriptions for the tags? None of them apply to your problem – Andrei Dec 14 '19 at 17:02
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$$2^{(a+1)!}(1+\frac{1}{2^{1}}+\frac{1}{2^{2}}+...)=2^{(a + 1)!} \left ( \sum_{i = 0}^\infty 0.5^i \right ) $$
$$ = 2^{(a + 1)!} 2$$
$$ = 2^{(a + 1) a!} 2$$
$$ = \left ( 2^{a!} \right )^{a+1} 2 $$
ironX
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