Prove: $\frac{1}{4(\ln2)^2}\leq\sum_{n=0}^{\infty}\frac{2^n}{2^{2^n}}$ I'd like your help with proving that $\frac{1}{4(\ln2)^2}\leq\sum_{n=0}^{\infty}\frac{2^n}{2^{2^n}}$.
I don't really know where to start. I tried to find a power series to integrate or derive in order to reach something close but it's just impossible. 
Any suggestion?
Edit:
Thanks, but I'm looking for a proof which uses calculus theorems and methods, and not to take the first two numbers of the series or something like that.
Thanks!
 A: For the begining we prove that $f(n)=\frac{2^n}{2^{2^n}}$ is strictly monotone decreasing. Indeed, $f(x)=2^{x-2^x}$ and for all $x>0$ we have $f'(x)=2^{x-2^x}(1-2^x\ln2)<0$. Hence $f$ is decreasing and we can make estimation:
$$
\sum\limits_{n=0}^\infty f(n)=
\sum\limits_{n=0}^\infty\int\limits_{[n,n+1]}f(n)dx=
\sum\limits_{n=0}^\infty\int\limits_{[n,n+1]}f([x])dx\geq
$$
$$
\sum\limits_{n=0}^\infty\int\limits_{[n,n+1]}f(x)dx=
\int\limits_{[0,+\infty)} f(x)dx
$$
Note that
$$
\int\limits_{[0,+\infty)} f(x)dx=
\int\limits_{[0,+\infty)} \frac{2^x}{2^{2^x}}dx=
\int\limits_{[0,+\infty)} \frac{d(2^x)}{2^{2^x}\ln2}=
\int\limits_{[1,+\infty)} \frac{du}{2^u\ln2}=
$$
$$
\frac{1}{\ln2}\left(-\frac{2^{-u}}{\ln2}\right)_{u=1}^{u=+\infty}=\frac{1}{2\ln^2 2}>\frac{1}{4\ln^2 2}
$$
A: By applying the Cauchy condensation test to $\sum_{n=1}^\infty \frac{1}{2^n}$ you get
$$\sum_{n=1}^\infty \frac{1}{2^n} \leq \sum_{n=0}^{\infty}\frac{2^n}{2^{2^n}} \leq 2\sum_{n=1}^\infty \frac{1}{2^n}  \,.$$
Calculating the geometric series $\sum_{n=1}^\infty \frac{1}{2^n}=1$ you get
$$\sum_{n=0}^{\infty}\frac{2^n}{2^{2^n}} \geq 1 = \frac{1}{(\ln e)^2}\geq \frac{1}{ (\ln(4))^2} =\frac{1}{4(\ln2)^2}$$
P.S. This is basically David Mitra's solution, complicated a lot using Calculus..
