Can we get an approximate lower bounds for the following expression:
$$\left( 1 + \frac{1}{ 2 \left( \frac{4^{nC}-1}{2^n-1} \right) } \right)^{ \frac{1}{\left( \frac{4^{nC}-2^{nC}}{2^n-1} \right) } } -1$$
...where $n$ and $C$ are positive integers from around 10 to 1000?
Or, alternatively, and better, can we get an upper bounds for the reciprocal of this function?
Your expression is $\left( 1 + \frac{1}{ 2 \left( \frac{4^{nC}-1}{2^n-1} \right) } \right)^{ \frac{1}{\left( \frac{4^{nC}-2^{nC}}{2^n-1} \right) } } -1$.
Let $g(n) = \frac{4^{nC}-1}{2^n-1} $ and $h(n) = \frac{4^{nC}-2^{nC}}{2^n-1} $. You want $f(n) =(1+\frac1{2g(n)})^{1/h(n)} $ (ignoring the "$-1$" for now).
Since $C$ is moderately large, $g(n) \approx \frac{4^{nC}}{2^n} =2^{2nC-n} =2^{n(2C-1)} $ and $h(n) \approx \dfrac{4^{nC}}{2^n} = 2^{2nC-n} = 2^{n(2C-1)} $.
Letting $D = 2C-1$ and $a(n) =2^{Dn} $, $g(n) \approx a(n) $ and $h(n) \approx a(n) $ so $f(n) \approx (1+\frac1{2a(n)})^{1/a(n)} $.
Looking at this, I find myself asking if you want $a(n)$, not $1/a(n)$, in the exponent. If this were the case, the limit would be $e^{1/2}$.
Similarly, if you want $a(n)$, not $1/a(n)$, in the internal expression, the limit would be $1$ (since $m^{1/m} \to 1$).
In your case, $(1+\frac1{2a(n)})^{1/a(n)} =e^{\ln(1+\frac1{2a(n)})/a(n)} \approx e^{(1/(2a(n))/a(n)} = e^{1/(2a^2(n))} \to 1+1/(2a^2(n)) $.
Since you have that "$-1$" in your expression, subtract $1$ from these results.
(added later)
Since $e^x \approx 1+x$ for small $x$ and $a(n) = 2^{Dn}$, $e^{1/(2a^2(n))}-1 \approx 1/(2a^2(n)) =1/(2\ 2^{2Dn}) =1/2^{2Dn+1} =1/2^{2(2C-1)n+1} $
