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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 $A_n$?

barak manos
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    Use the Lambert function and solve for $A_n$. –  Aug 06 '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 06 '14 at 12:02
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    $$\sum_{n=1}^\infty \frac{1}{2^n\log(2^n)} = \frac{1}{\log 2}\sum_{n=1}^\infty \frac{1}{n}\left(\frac12\right)^n = - \frac{1}{\log 2}\log\left(1-\frac12\right) = 1$$ – achille hui Aug 06 '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 $A_n=\frac{2^n}{W(2^n)}$. –  Aug 06 '14 at 12:16
  • @ioannisgalidakis Cheers for responding and clarifying :) – Jam Aug 06 '14 at 12:21
  • @achillehui: That doesn't quite work out for me, because $\sum\limits_{n=1}^{\infty}\frac{1}{2^n}=1$. So are you suggesting that $\sum\limits_{n=1}^{\infty}\frac{1}{2^n}=\sum\limits_{n=1}^{\infty}\frac{1}{2^nln(2^n)}$? – barak manos Aug 07 '14 at 15:03
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    @barakmanos You can ask WA to evaluate that in closed form. – achille hui Aug 07 '14 at 15:34
  • @achillehui: Nice! You just made me submit another question here (http://math.stackexchange.com/q/890479/131263). Add it as an answer and I will grant you the V. Thanks :) – barak manos Aug 07 '14 at 21:56

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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 $\displaystyle\;\sum_{n=1}^\infty\frac{1}{A_n\log A_n} = \sum_{n=1}^\infty\frac{1}{2^n\log(2^n)} = 1$.

achille hui
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