I'm trying to find distribution on natural numbers with maximum entropy, such that its expectation is equals 2 with Lagrange method. My problem is that i can not solve system of equations
$$ L(\lambda_1, \lambda_2, p_1, p_2, ..., p_n, ...) = -\sum p_n\log(p_n) + \lambda_1(\sum np_n - 2) + \lambda_2(\sum p_n - 1)$$
It's something like $$ L'_{p_n} = -\log(p_n) - 1 + \lambda_1n + \lambda_2 $$ $$L'_{\lambda_1} = \sum np_n - 2$$ $$L'_{\lambda_2} = \sum p_n - 1$$
and then
$$p_n = e^{\lambda_1n + \lambda_2 - 1} \\ \sum ne^{\lambda_1n + \lambda_2 - 1} = 2 \\ \sum e^{\lambda_1n + \lambda_2 - 1} = 1$$
just evaluate the summations to get relations between $\lambda_1$ and $\lambda_2$
$\sum e^{\lambda_1 n + \lambda_2 - 1} = 1\\ e^{\lambda_2 - 1}\sum e^{\lambda_1 n} = 1 \\ e^{\lambda_2 - 1} \frac{e^{\lambda_1}}{1 - e^{\lambda_1}} = 1 \rightarrow (1)$
for other summation, we can use the fact that $\sum_{n = 1}^\infty n*r^n = \frac{r}{(1-r)^2}$, when $|r| < 1$
$\sum ne^{\lambda_1 n + \lambda_2 - 1} = 2\\ e^{\lambda_2 - 1} \sum ne^{\lambda_1 n} = 2\\ e^{\lambda_2 - 1} \frac{e^{\lambda_1}}{(1 - e^{\lambda_1})^2} = 2$
using (1) we get $\frac{1}{1 - e^{\lambda_1}} = 2\\ \frac{1}{2} = 1 - e^{\lambda_1} \\ e^{\lambda_1} = \frac{1}{2} \\ \lambda_1 = -ln(2)$
substituting in (1) we get $e^{\lambda_2 - 1} = 1 \implies \lambda_2 = 1$
so $p_n = \frac{1}{2^n}$ and the entropy is $-\sum \frac{1}{2^n} log(\frac{1}{2^n}) = log(4) \approx 1.38629$