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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!

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    $\begingroup$ The left hand side is less than 1. The right hand side is greater than 1 (take the first two terms of the series). $\endgroup$
    – David Mitra
    Commented Feb 3, 2012 at 13:53
  • $\begingroup$ The left hand side is about 0.52, the terms of the right hand side are all positive, and the first two are $\frac{2^0}{2^{2^0}}=\frac{1}{2}$ and $\frac{2^1}{2^{2^1}}=\frac{1}{2}$ $\endgroup$
    – Thomas Andrews
    Commented Feb 3, 2012 at 13:53
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    $\begingroup$ If you want a proof that $\frac{1}{4(\log 2)^2}$ is less then 1, you can just show that $2\log 2=\log 4 > 1$. But $\log 4 = \int_1^4 \frac{dx}{x}>\frac{1}{2} + \frac{1}{3}+ \frac{1}{4}> 1$ $\endgroup$
    – Thomas Andrews
    Commented Feb 3, 2012 at 13:57
  • $\begingroup$ 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. $\endgroup$
    – Jozef
    Commented Feb 3, 2012 at 14:00
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    $\begingroup$ "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." - That seems entirely unreasonable to me. So you want a proof using calculus, but you reject proofs using basic/elementary arithmetic and algebra? You can't even have calculus until you accept that x < y if x < 1 and y >= 1. $\endgroup$
    – Patrick87
    Commented Feb 3, 2012 at 18:48

2 Answers 2

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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..

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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 $$ \int\limits_{[0,+\infty)} f(x)dx= \int\limits_{[0,+\infty)} \frac{2^x}{2^{2^x}}dx= \int\limits_{[0,+\infty)} \frac{d(2^x)}{2^{2^x}\ln2}= \int\limits_{[1,+\infty)} \frac{du}{2^u\ln2}= $$ $$ \frac{1}{\ln2}\left(-\frac{2^{-u}}{\ln2}\right)_{u=1}^{u=+\infty}=\frac{1}{2\ln^2 2}>\frac{1}{4\ln^2 2} $$

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  • $\begingroup$ Thank you. @Norbert. Why is this inequality correct? $\endgroup$
    – Jozef
    Commented Feb 3, 2012 at 15:11
  • $\begingroup$ Or note that $\sum_{n=0}^\infty f(n) = \int_0^{+\infty} f([x])dx \leq \int_{0}^{+\infty} f(x) dx$, where $[x]$ is the largest integer $\leq x$, and the inequality follows from $f([x])\leq f(x)$ since $f$ is monotone. $\endgroup$
    – Thomas Andrews
    Commented Feb 3, 2012 at 15:39
  • $\begingroup$ Thank you guys. so @Norbert: do you consider the area left to the graph as negative? $\endgroup$
    – Jozef
    Commented Feb 3, 2012 at 16:34
  • $\begingroup$ @Jozef I found some mistakes, so see my edits. $\endgroup$
    – Norbert
    Commented Feb 3, 2012 at 16:58

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