Can a TI89 or other CAS calculators solve this?
I tried it on my classpad 330 did not work
solve for p
$$p = \sum_{j=0}^\infty p^j \frac{2^je^{-2}}{j!}$$
solve for p
$$p = \lim_{z \to \infty} \sum_{j=0}^z p^j \frac{2^je^{-2}}{j!}$$
Edit: The whole point of resorting to a calculator is if I would not be clever enough to see that I can rewrite it as @MPW suggested. This is what my textbook does as well. Once you see that, you don't need a CAS calculator to solve the rest...
As MPW wrote, the right hand side is $e^{2p-2}$. So, you want to solve the equation $$p=e^{2p-2}$$ Any equation which can write (after possible rearrangements as $$A+Bx+C\log(Dx+E)=0$$ has a solution which can be expressed in terms of Lambert function.
For the specific case of your post, the solution is (beside the trivial $p=1$) $$p=-\frac{1}{2} W\left(-\frac{2}{e^2}\right) \simeq 0.2031878699799799538384791$$ You can also look at this problem considering the intersection of $y=\log(p)$ and $y=2p-2$. You would notice that the solution is around $p_0=\frac{1}{4}$ and start Newton iteration $$p_{n+1}=p_n-\frac{f(p_n)}{f'(p_n)}$$ The successive iterates will then be $0.193147$, $0.202777$, $ 0.203187$, $0.203188$ which is the solution for six significant figures.
