Prove $\sum_{n=0}^\infty\frac{2^n}{2^{2^n}}\geq\frac{1}{4(\ln 2)^2}$

Question

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?

Answer 1

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 $f(0)=1/2>0$.

Answer 2

$$\sum_{n=0}^\infty\frac{2^n}{2^{2^n}}>\int\limits_{1}^{+\infty}\frac{2^x}{2^{2^x}}dx=-\frac{1}{2^{2^x}\ln^2 2}|_1^{+\infty}=\frac{1}{4\ln^22}$$