<s>[INST] $\ln $ and Taylor Series Expansion (what went wrong) Edited
Problem
I'm trying to express $\ln{(1-(\frac{N}{K})^{\frac{1}{4}})}$ in terms of $\ln N$, where $K$ is a constant and $1 \leq N \leq K$. This also implies $\frac{N}{K} \leq 1$. 

Anyone knows if it is possible to represent $\ln{(1-(\frac{N}{K})^{\frac{1}{4}})}$ in terms of $\ln N$ ?
  Cuz as seen below, using the Taylor Expansion just returns the original term back to us and it didn't seem to help a lot.  

Taylor Series Expansion Used
The Taylor series I'll be using is: (refer to this Website for Taylor Series Expansions)
$$\ln (1+x) = x- \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots $$
where $-1 \leq x \leq 1$. 
Solving
Then, observe that $\ln{(1-(\frac{N}{K})^{\frac{1}{4}})} = \ln{(1-(\frac{N}{K})^{\frac{1}{8}}) (1+(\frac{N}{K})^{\frac{1}{8}})} = \ln(1-\frac{N}{K})^{\frac{1}{8}} + \ln (1+\frac{N}{K})^{\frac{1}{8}}$. 
By the series expansion: we get: 
$$\ln(1-\frac{N}{K})^{\frac{1}{8}} = -(\frac{N}{K})^{\frac{1}{8}}-\frac{(\frac{N}{K})^\frac{1}{4}}{2} - \frac{(\frac{N}{K})^\frac{3}{8}}{3}- \frac{(\frac{N}{K})^\frac{1}{2}}{4} - \ldots$$
and
$$\ln(1+\frac{N}{K})^{\frac{1}{8}} = (\frac{N}{K})^{\frac{1}{8}}-\frac{(\frac{N}{K})^\frac{1}{4}}{2} + \frac{(\frac{N}{K})^\frac{3}{8}}{3}- \frac{(\frac{N}{K})^\frac{1}{2}}{4} - \ldots$$
Thus taking the sum of the 2 logarithms above, we get:
$$\ln{(1-(\frac{N}{K})^{\frac{1}{4}})} =- (\frac{N}{K})^\frac{1}{4} - \frac{(\frac{N}{K})^\frac{1}{2}}{2} - \frac{(\frac{N}{K})^\frac{3}{4}}{3} - \ldots $$

Notice that if we were to differentiate the series, we get: 
$\frac{d}{dx} x + \frac{1}{2}(x^2) + \frac{1}{3}(x^3) + \ldots = 1+x+x^2+ \ldots = \frac{1}{1-x}$, for $|x| \leq 1$. 
Hence, $x + \frac{1}{2}(x^2) + \frac{1}{3}(x^3) + \ldots = \int \frac{1}{1-x} dx = -\ln{(1-x)} + C$  for some arbitrary constant $C$. 

Hence, $\ln{(1-(\frac{N}{K})^{\frac{1}{4}})} = -[(\frac{N}{K})^\frac{1}{4} + \frac{(\frac{N}{K})^\frac{1}{2}}{2} + \frac{(\frac{N}{K})^\frac{3}{4}}{3} - \ldots]  = \int -\frac{1}{1-(\frac{N}{K})^\frac{1}{4}} d\frac{N}{K} = \ln {(1-(\frac{N}{K})^\frac{1}{4})} + C$.
We get back the same term!
 [/INST] Note that instead of going through the steps of adding the
Taylor series for $\ln\left(1+\left(\frac NK\right)^{1/8}\right)$
and $\ln\left(1-\left(\frac NK\right)^{1/8}\right)$,
you could have derived the series
$$\ln\left(1-\left(\frac NK\right)^{\frac14}\right) 
=  -\left(\frac NK\right)^\frac14
  - \frac{\left(\frac NK\right)^\frac12}{2}
  - \frac{\left(\frac NK\right)^\frac34}{3} - \ldots $$
simply by applying the known Taylor series expansion of $\ln(1+x)$
with $x = -\left(\frac NK\right)^{1/4}.$
Your method of differentiating and then integrating the series
is a partial confirmation that the series is the 
correct Taylor series of $\ln\left(1-\left(\frac NK\right)^{1/4}\right).$
(You would completely confirm the correctness of the series
if you showed that $C=0$.)
So it's hardly surprising that this manipulation of the series
gives you back what you started with (plus the pesky unknown constant $C$).
Of course you conceivably could have started with a much more complicated
expression that happened to give the same Taylor series and therefore
was equal to $\ln\left(1-\left(\frac NK\right)^{1/4}\right).$
Getting back the same expression is a clue that perhaps the expression
$\ln\left(1-\left(\frac NK\right)^{1/4}\right)$ is already about as
simple as it should be.
There's no reason to expect $\ln N$ to show up anywhere in any of these
calculations unless you force it to occur by substituting $N = e^{\ln N}$.
In fact,
$$\ln\left(1-\left(\frac NK\right)^{\frac14}\right)
= \ln\left(1-e^{(\ln N - \ln K)/4}\right),$$
so unless you have a nicer way of writing $\ln\left(1-e^{(x - \ln K)/4}\right)$
in terms of $x$ then I think this may be as good as it gets.
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