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I have been trying to simplify the following summation with the intention of breaking it into less complex summations, but I keep getting stuck no matter what I try:

$$\sum_{i=0}^{k-1} 3^{i} \cdot \frac{\sqrt{\frac{n}{3^i}}}{\log_{2}\frac{n}{3^i}}$$

Among the things I tried was raising to the power of $1/2$* the upper part of the fraction to cancel out the $3^i$, but after that I'm left with $\sqrt{n\,3^i}$ anyways without any clear steps to simplify further. Also tried putting the lower part of the fraction as a difference of logarithms, but got stuck in the same way. I even tried to use the change of base identity to put the lower part of the fraction in terms of $\log_3(2)$ and see if I could have canceled something but no luck.

Just simplifying the sum would be good enough since I can try to take it from there by substituting $k\; (k = \log_3(n)$ in case that is useful).

Noa Even
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MikeKatz45
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  • May be this is simpler (assuming the Algebra is correct :)): $$\sqrt{n} \cdot \sum_{i=0}^{k-1} \frac{\sqrt{3^{i}} }{p-i \cdot k}$$ where $k=log_{2}{3}, p=\log_{2}{n}.$ Note that both p, k are constants. – NoChance Jan 28 '19 at 11:25
  • it is kinda simplier yeah, but the idea is to basically put the 2 terms that depend on the index in separate sums somehow. That way they could be solved using identities. – MikeKatz45 Jan 28 '19 at 19:36
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    I don't see any reason to think it's possible to do better than what @NoChance has done. But NoChance has been careless in using $k$ to stand for $\log_23$, when $k$ is already being used in the upper limit of the summation. – Gerry Myerson Mar 14 '19 at 01:04
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    @GerryMyerson thank you for your comment. You are correct. – NoChance Mar 14 '19 at 08:33

2 Answers2

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The issue expression can be presented in the form of $$S={\large\sum\limits_{i=0}^{k-1}}3^i\dfrac{\sqrt{\dfrac n{3^i}}}{\log_2\dfrac{n}{3^i}} = \sqrt n \log_32{\large\sum\limits_{i=0}^{k-1}}\dfrac{(\sqrt3)^i}{\log_3{n}-i},$$ so $$S=\sqrt n \log_32\left(3^{k/2}\Phi(\sqrt3,1,k-\log_3n)-\Phi(\sqrt3,1,-\log_3n)\right)$$ (see also Wolfram Alpha), where

$$\Phi(z,s,a) = {\large\sum\limits_{m=0}^\infty\,\dfrac {z^m}{(m+a)^s}}$$ is Lerch transcendent.

Assuming $k=\lceil\log_3n -1\rceil,$ can be built the plots for the issue sums

The issue sums

and via Lerch transcendent

Lerch transcendent sums

Easy to see that summation via Lerch transcendent is correct.

Yuri Negometyanov
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If I plug $$ \sum_{i=0}^{\log_3(n)-1}3^i \frac{\sqrt{\frac{n}{3^i}}}{\log_2(\frac{n}{3^i})} $$ into Wolfram Mathematica, then I get no simplification, so an nice simplification probably does not exist. The best we can probably do is, with thanks to @NoChance, $$ \sum_{i=0}^{\log_3(n)-1}3^i \frac{\sqrt{\frac{n}{3^i}}}{\log_2(\frac{n}{3^i})} = \sum_{i=0}^{\log_3(n)-1}3^i \frac{\sqrt{\frac{n}{3^i}}}{\log_2(n)-\log_2(3^i))} \\ =\sum_{i=0}^{\log_3(n)-1} 3^i\frac{\sqrt{\frac{n}{3^i}}}{\log_2(n)-i\log_2(3))} \\ =\sum_{i=0}^{\log_3(n)-1} \sqrt{n}\frac{3^i\sqrt{\frac{1}{3^i}}}{\log_2(n)-i\log_2(3))} \\ = \sqrt{n}\sum_{i=0}^{\log_3(n)-1}\frac{\sqrt{3^i}}{\log_2(n)-i\log_2(3))} \\ $$ If I plug this into Wolfram Mathematica, then we see that we get something $$ \sqrt{n}\sum_{i=0}^{\log_3(n)-1}\frac{\sqrt{3^i}}{\log_2(n)-i\log_2(3))} = -\frac{n\bigg(\log(3-\sqrt{3})-\log(3)\bigg)}{\log_2(3)}-\frac{\sqrt{n}HyperGeometric2F1\bigg(1,\log_3(3n),2+\log_3(n),\frac{1}{\sqrt{3}}\bigg)}{\sqrt{3}\log_2{3n}} $$ If we look at Wikipedia, https://en.wikipedia.org/wiki/Hypergeometric_function, for some more info on the Hypergeometric2F1 we see that there are no further simplifications listed.

ChocolateRain
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