<s>[INST] Prove $\sum_{n=0}^\infty\frac{2^n}{2^{2^n}}\geq\frac{1}{4(\ln 2)^2}$ 
Prove that:
  $$\sum_{n=0}^\infty\frac{2^n}{2^{2^n}}\geq\frac{1}{4(\ln 2)^2}$$

I computed the indefinite integral:
$$\int\frac{2^x}{2^{2^x}}dx=-\frac{1}{2^{2^x}(\ln 2)^2}+C$$
How can I continue from here?
 [/INST] You are on the right track. Note that $2^x-x$ is increasing for $x\geq 1$. Therefore $f(x)=\frac{1}{2^{2^x-x}}$ is decreasing and
$$\sum_{n=1}^\infty\frac{2^n}{2^{2^n}}=\sum_{n=1}^\infty f(n)\geq \sum_{n=1}^\infty \int_{n}^{n+1}f(x) dx=\int_1^{\infty}f(x)dx=\left[-\frac{1}{2^{2^x}(\ln 2)^2}\right]_1^{\infty}=\frac{1}{4(\ln 2)^2}.$$
This is a stronger inequality because the sum starts form the index $1$ and $f(0)=1/2>0$.
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