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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).
['https://math.stackexchange.com/questions/3090509', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/635189/']
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](https://www.wolframalpha.com/input/?i=sum_(i%3D0,k-1)%20a%5Ei%2F(b-i))), where

> 
> $$\Phi(z,s,a) = {\large\sum\limits\_{m=0}^\infty\,\dfrac
> {z^m}{(m+a)^s}}$$
>  is [Lerch transcendent](http://mathworld.wolfram.com/LerchTranscendent.html).
> 
> 
> 

Assuming $k=\lceil\log\_3n -1\rceil,$ can be built the plots [for the issue sums](https://www.wolframalpha.com/input/?i=plot%20table%20sum_(j%3D0,%20ceil(log_3(n)%20-2))%20(sqrt(n)%20%20log_3(2)%203.0%5E(j%2F2)%20%2F(log_3(n)-j)),%20%7Bn,3,99%7D)

[![The issue sums](https://i.stack.imgur.com/oUqlQ.png)](https://i.stack.imgur.com/oUqlQ.png)

and [via Lerch transcendent](https://www.wolframalpha.com/input/?i=plot%20%7Bsqrt(n)%20%20log_3(2)%20(3%5E(ceil(log_3%20n%20-1)%2F2)%20HurwitzLerchPhi(sqrt3,%201,%20ceil(log_3%20n%20-1)%20-%20log_3(n))%20-%20HurwitzLerchPhi(sqrt3,%201,%20%20-%20log_3%20(n)))%7D,%20n%3D3,%2099)

[![Lerch transcendent sums](https://i.stack.imgur.com/3jXyV.png)](https://i.stack.imgur.com/3jXyV.png)

Easy to see that summation via Lerch transcendent is correct.
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.