How to calculate entropy of output of a PRNG Suppose the entropy of seed key is $H(n)$ which is $n$-bit long and random and the output bit length of the PRNG is $2^n$, which will be used as a long running key. Though it is obvious in that case, as output key sequence of PRNG is derived from input seed key, so entropy of output key sequence of PRNG never exceeds the entropy of the seed key. $H(2^n) \le H(n)$ in this scenario but how to prove it?
TL;DR: The applied thought process was wrong. Entropy is a function of random variables and not of arbitrary integers. The fact that the PRNG doesn't increase entropy follows from the fact that the input space is limited and there's no other input to the algorithm and thereby entropy can't be increased.



First, let's recall the formal definition of "entropy" (as can be found in the Handbook of Applied Cryptography, chapter 2.2, PDF):


  Definition The entropy or uncertainity of $X$ is defined to be $H(X)=-\sum_{i=1}^np_i\lg p_i=\sum_{i=1}^np_i\lg (\frac{1}{p_i})$ where, by convention, $p_i\cdot \lg p_i=p_i\cdot \lg (\frac{1}{p_i})=0$ if $p_i=0$.


Notation: $\lg a$ is the binary logarithm of $a$. $X$ is a random variable that takes each value $x_i$ (from a finite set if size $n$) with the correspdoning probability $p_i$.



Now let's model the $m$-bit string as the random variable $X$. The $m$-bit string can take $n_X=2^m$ values, each of which has the probability $p_i=\frac{1}{n_X}=2^{-m}$. Thereby the entropy $H(X)$ is $$H(X)=\sum_{i=1}^np_i\lg (\frac{1}{p_i})=\sum_{i=1}^{2^m}2^{-m}\lg (2^m)=2^m\cdot(2^{-m}m)=m$$ as expected.



The next step is to properly model the "PRNG". I will model it as a pseudo random generator (PRG), because I assume that's how you meant it.


  In short: A pseudo random generator (PRG) is a deterministic algorithm that takes a fixed length input and returns a longer output.


(not a full formal definition and the usual definition would restrict to polynomially sized outputs, but we don't need one here)

By this we can model our PRG $\mathcal G: \{0,1\}^m\rightarrow \{0,1\}^{2^m}$. As $\{0,1\}^m$ can only take $2^m$ states and this is the only input to the function, we know that there can be at most $2^m$ different outputs of the PRG. 

If we model the PRG output as a random variable $Y$, we know that (in "best case") $p_i=2^{-m}$ for $2^m$ different $y_i$ and $n_Y=2^{2^m}$. We further know that for the remaining $2^{2^{m}}-2^m$ possible values $y_i$, $p_i=0$ holds. With this information, we can compute the upper bound of the entropy of the PRG:

$$H(Y)=\sum_{i=1}^np_i\lg (\frac{1}{p_i})=\sum_{i=1}^{2^m}2^{-m}\lg (2^m)+\sum^{2^{2^m}-2^m}_{i=1}0\cdot\lg(\frac{1}{0})=2^m\cdot(2^{-m}m)+(2^{2^m}-2^m)\cdot 0=m$$

Note that $0\cdot \lg(\frac{1}{0})$ is, by convention, $0$.



As we have modeled $Y$ using as an optimal "PRG", it may be very well the case that there aren't $2^m$ different outputs. So we can conclude with $$H(X)\geq H(Y)$$ as expected and thereby that the PR(N)G doesn't increase entropy by itself.