Simplifying the summation $(3^i\sqrt{n/3^i})/\log_2(n/3^i)$ from $i=0 \text{ to }k-1$

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).

• 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 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 at 19:36
• 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 at 1:04
• @GerryMyerson thank you for your comment. You are correct. – NoChance Mar 14 at 8:33

2 Answers

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.

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

Easy to see that summation via Lerch transcendent is correct.