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Some Double-Exponential Series

2017 Jan 7

Evaluate the following infinite series: $$\sum_{n=0}^\infty \frac{2^n}{2^{2^n}+1}$$ $$\sum_{n=0}^\infty \frac{2^{-n}}{2^{2^{-n}}+1}$$ $$\sum_{n=0}^\infty \frac{4^{n+2^n}}{(4^{2^n}+1)^2}$$

All of these series are telescoping series, but determining how to telescope them can be difficult and involve a lot of nasty algebra, so I shall instead show how to derive their values from infinite products that are easy to evaluate using logarithmic differentiation. Consider first the infinite product $$\prod_{n=0}^\infty (1+\alpha^{2^n})$$ where $\alpha$, with $|\alpha|\lt 1$, is a parameter. We are leaving $\alpha$ as a variable for now so that we may differentiate with respect to it later. This is an easy product to evaluate if one recalls the following "difference of squares" factoring trick from algebra: $$x^2-y^2=(x-y)(x+y)$$ This trick can be extended to greater heights: $$x^{2^n}-y^{2^n}=(x-y)(x+y)(x^2+y^2)...(x^{2^{n-1}}+y^{2^{n-1}})$$ We shall use this trick to telescope this product into a closed form: $$\begin{align} \prod_{n=0}^\infty (1+\alpha^{2^n})&= \lim_{m\to\infty }\prod_{n=0}^m (1+\alpha^{2^n})\\ &= \frac{1}{1-\alpha}\lim_{m\to\infty } (1-\alpha)\prod_{n=0}^m (1+\alpha^{2^n})\\ &= \frac{1}{1-\alpha}\lim_{m\to\infty } (1-\alpha^{2^{m+1}})\\ &= \frac{1}{1-\alpha} \end{align}$$ which gives us the result $$\prod_{n=0}^\infty (1+\alpha^{2^n})=\frac{1}{1-\alpha}$$ Now we shall apply logarithmic differentiation - that is, we will take the logarithm of both sides, and then differentiate. Taking the logarithm, we have $$\ln \prod_{n=0}^\infty (1+\alpha^{2^n})=-\ln(1-\alpha)$$ $$\sum_{n=0}^\infty \ln(1+\alpha^{2^n})=-\ln(1-\alpha)$$ and differentiating both sides with respect to $\alpha$, we have $$\sum_{n=0}^\infty \frac{2^n\alpha^{2^{n}-1}}{1+\alpha^{2^n}}=\frac{1}{1-\alpha}$$ or $$\sum_{n=0}^\infty \frac{2^n}{1+\alpha^{-2^n}}=\frac{\alpha}{1-\alpha}$$ By taking $\alpha=1/2$, we obtain the value of the first series: $$\bbox[5px,border:2px solid gray] { \sum_{n=0}^\infty \frac{2^n}{1+2^{2^n}}=1 }$$ We may also take the previous result and differentiate again with respect to $\alpha$ for another formula: $$\sum_{n=0}^\infty \frac{4^n\alpha^{2^n}}{(1+\alpha^{2^n})^2}=\frac{\alpha}{(1-\alpha)^2}$$ Letting $\alpha=4$ gives the solution to the third problem: $$\bbox[5px,border:2px solid gray] { \sum_{n=0}^\infty \frac{4^{n+2^n}}{(1+4^{2^n})^2}=\frac{4}{9} }$$ If we rearrange this last formula a little bit, we have $$\sum_{n=0}^\infty \frac{4^n\alpha^{2^n}}{(1+\alpha^{2^n})^2}=\frac{\alpha}{(1-\alpha)^2}$$ $$\sum_{n=0}^\infty \frac{4^n}{(\alpha^{-2^{n-1}}+\alpha^{-2^{n-1}})^2}=\frac{\alpha}{(1-\alpha)^2}$$ $$\sum_{n=0}^\infty \frac{4^n}{4\cosh^2(2^{n-1}\ln\alpha)}=\frac{\alpha}{(1-\alpha)^2}$$ $$\sum_{n=0}^\infty \frac{4^{n-1}}{\cosh^2(2^{n-1}\ln\alpha)}=\frac{\alpha}{(1-\alpha)^2}$$ $$\sum_{n=0}^\infty \frac{4^{n}}{\cosh^2(2^{n}\ln\alpha)}=\frac{\alpha}{(1-\alpha)^2}-\frac{1}{4\cosh^2(\frac{\ln(\alpha)}{2})}$$ After a lot of gross algebra and some properties of the hyperbolic functions, this turns into $$\sum_{n=0}^\infty \frac{4^{n}}{\cosh^2(2^{n}\ln\alpha)}=\frac{1}{\cosh^2(\ln\alpha)-1}$$ or, letting $\ln\alpha=\beta$, $$\bbox[lightgray,5px] { \sum_{n=0}^\infty \frac{4^{n}}{\cosh^2(2^{n}\beta)}=\frac{1}{\cosh^2(\beta)-1} }$$ This is our polished formula, which will later be accompanied by a similar (but more difficult to prove) twin formula.

Another product can be obtained using the "difference of cubes" factoring trick $$x^3-y^3=(x-y)(x^2+xy+y^2)$$ and its generalization: $$x^{3^n}-y^{3^n}=(x-y)(x^2+xy+y^2)(x^6+x^3y^3+y^6)...(x^{2\cdot 3^{n-1}}+x^{3^{n-1}}y^{3^{n-1}}+y^{2\cdot 3^{n-1}})$$ From this trick, we have that if $|\alpha|\lt 1$, $$\begin{align} \prod_{n=0}^\infty (1+\alpha^{3^n}+\alpha^{2\cdot 3^n})&= \lim_{m\to\infty }\prod_{n=0}^m (1+\alpha^{3^n}+\alpha^{2\cdot 3^n})\\ &= \frac{1}{1-\alpha}\lim_{m\to\infty } (1-\alpha)\prod_{n=0}^m (1+\alpha^{3^n}+\alpha^{2\cdot 3^n})\\ &= \frac{1}{1-\alpha}\lim_{m\to\infty } (1-\alpha^{3^{m+1}})\\ &= \frac{1}{1-\alpha} \end{align}$$ Of course, these tricks can be taken even further using formulas for differences of fourth, fifth, and higher powers, but this is as far as I will take it.

Now consider the product $$\prod_{n=0}^\infty \frac{1+\alpha^{2^{-n}}}{2}$$ with $\alpha\gt 0$. Using the difference of squares factoring trick, we have $$\begin{align} \prod_{n=0}^\infty \frac{1+\alpha^{2^{-n}}}{2}&= \lim_{m\to\infty} \prod_{n=0}^m \frac{1+\alpha^{2^{-n}}}{2}\\ &= \lim_{m\to\infty} \frac{1-\alpha^{2^{-m}}}{2^{m+1} (1-\alpha^{2^{-m}})}\prod_{n=0}^m (1+\alpha^{2^{-n}})\\ &= \lim_{m\to\infty} \frac{1}{2^{m+1} (1-\alpha^{2^{-m}})}\cdot (1-\alpha^{2^{-m}})\prod_{n=0}^m (1+\alpha^{2^{-n}})\\ &= \lim_{m\to\infty} \frac{1}{2^{m+1} (1-\alpha^{2^{-m}})}\cdot (1-\alpha^2)\\ &= (1-\alpha^2)\lim_{m\to\infty} \frac{2^{-m-1}}{1-\alpha^{2^{-m}}}\\ &= (1-\alpha^2)\lim_{m\to\infty} \frac{2^{-m-1}\cdot -\ln(2)}{-\alpha^{2^{-m}}\cdot \ln(\alpha)\cdot 2^{-m}\cdot -\ln(2)}\\ &= \frac{\alpha^2 -1}{2\ln \alpha} \end{align}$$ yielding the result $$\prod_{n=0}^\infty \frac{1+\alpha^{2^{-n}}}{2}=\frac{\alpha^2 -1}{2\ln \alpha}$$ By taking a logarithm of both sides, we have $$\ln \prod_{n=0}^\infty \frac{1+\alpha^{2^{-n}}}{2}=\ln|1-\alpha^2|-\ln|2\ln(\alpha)|$$ $$\sum_{n=0}^\infty \ln \frac{1+\alpha^{2^{-n}}}{2}=\ln|1-\alpha^2|-\ln|2\ln(\alpha)|$$ and differentiating gives us, after a bit of nasty algebra, $$\sum_{n=0}^\infty \frac{2^{-n}}{\alpha^{-2^{-n}}+1}=\frac{2x^2}{x^2-1}-\frac{1}{\ln(x)}$$ Using this formula with $\alpha=1/2$, we can obtain the value of the first sum: $$\bbox[5px,border:2px solid gray] { \sum_{n=1}^\infty \frac{2^{-n}}{2^{2^{-n}}+1}=\frac{1}{\ln 2}-\frac{2}{3} }$$ Differentiating the formula again (and a bit of algebra) yields the formula $$\sum_{n=0}^\infty \frac{4^{-n}\alpha^{-2^{-n}}}{(\alpha^{-2^{-n}}+1)^2}=\frac{1}{\ln^2(\alpha)}-\frac{4\alpha^2}{(\alpha^2-1)^2}$$ and with a bit of algebraic manipulation, this turns into $$\sum_{n=0}^\infty \frac{4^{-n}}{\cosh^2(2^{-n}\ln\alpha)}=\frac{1}{\ln^2(\alpha)}-\frac{1}{\sinh^2(\ln\alpha)\cosh^2(\ln\alpha)}$$ and, letting $\beta=\ln\alpha$, this becomes the "twin formula" mentioned earlier: $$\bbox[lightgray,5px] { \sum_{n=0}^\infty \frac{4^{-n}}{\cosh^2(2^{-n}\beta)}=\frac{1}{\beta^2}-\frac{1}{\sinh^2(\beta)\cosh^2(\beta)} }$$ Again, the trick used to obtain this formula can be extended to use difference of cube, fourth power, and higher degree factoring tricks, but that's all I will do with it for now.