# Compute a sequence $A_n$ such that $\sum\limits_{n=1}^{\infty}\frac{1}{A_n\ln(A_n)}=1$

How can we compute a sequence $A_n$ of positive real numbers, such that $\displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{A_n\ln(A_n)}=1$?

One way I can think of, is by defining $A_n\ln(A_n)=2^n$, but how do we extract the formula for $A_n$?

• Use the Lambert function and solve for $A_n$. – Yiannis Galidakis Aug 6 '14 at 11:37
• @ioannisgalidakis Are you sure you can solve for $A_n$ with the Lambert function? I can't see a way to do it. – Jam Aug 6 '14 at 12:02
• $$\sum_{n=1}^\infty \frac{1}{2^n\log(2^n)} = \frac{1}{\log 2}\sum_{n=1}^\infty \frac{1}{n}\left(\frac12\right)^n = - \frac{1}{\log 2}\log\left(1-\frac12\right) = 1$$ – achille hui Aug 6 '14 at 12:03
• @oliveeuler: $A_n\cdot\ln(A_n)=2^n\Rightarrow\exp(A_n\cdot\ln(A_n))=\exp(2^n)\Rightarrow A_n^{A_n}=\exp(2^n)$. The equation $x^x=y$ has the solution $x=\frac{\ln(y)}{W(\ln(y))}$, with $y=\exp(2^n)$, so $A_n=\frac{2^n}{W(2^n)}$. – Yiannis Galidakis Aug 6 '14 at 12:16
• @barakmanos You can ask WA to evaluate that in closed form. – achille hui Aug 7 '14 at 15:34

The most ugly piece in the sum $\sum\limits_{n=1}^\infty \frac{1}{A_n\log A_n}$ is the $\log(\cdot)$ in the denominator. If one want to construct an ansatz for the problem, one should look for something that get rid of this $\log(\cdot)$. The simplest choice is $A_n = \alpha^n$ for some $\alpha > 1$.
As a function of $\alpha$, we have
$$\sum_{n=1}^\infty\frac{1}{A_n\log A_n} = \sum_{n=1}^\infty\frac{1}{\alpha^n\log(\alpha^n)} = \frac{1}{\log\alpha}\sum_{n=1}^\infty\frac{1}{n}\left(\frac{1}{\alpha}\right)^n\\ = -\frac{1}{\log\alpha}\log\left(1-\frac{1}{\alpha}\right) = 1 - \frac{\log(\alpha-1)}{\log \alpha}$$ If is then clear if we pick $\alpha = 2$, we will get $\displaystyle\;\sum_{n=1}^\infty\frac{1}{A_n\log A_n} = \sum_{n=1}^\infty\frac{1}{2^n\log(2^n)} = 1$.