The closed form solution is $P_j = 0, j \ge 0$
I shall show now that there is no solution to the recursive equations with
(1) $P_j \gt 0, j = 1, 2, 3, ...$
First from
$\alpha P_1 = P_2 + P_3$
we conclude
$\alpha \gt 0$
Notice also the $P_0$ appears only in the relation
$\alpha P_0 = P_1$
which shows that
$P_0 \gt 0$ as well, but $P_0$ does not appear in the normalization. Therefore we consider it as a mere abrevíation for $P_1/ \alpha $.
Now we transform the recursive relation into a standard form, which we define here to be one in which an element with a specific index is defined in terms of elements with smaller indices.
Define
(2) $Q_i = P_{i+1}+P_{i+2}+..., i = 0,1,2,...$
As a sum over positive quantities we have
$Q_i \gt 0, i = 0, 1, 2, ...$
The inversion of (2) is
(3) $P_i = Q_{i-1} - Q_i , i = 1, 2, ... $
Now the equations become
$\alpha P_j = Q_j - Q_{2j+1}, j =1, 2, 3, ... $
Using (3) we get
$\alpha (Q_{j-1}-Q_j) = Q_j - Q_{2j+1}$
or
q[2j+1] = (1+a) q[j] - a q[j-1], j = 1, 2, ... (4)
This is now a recursive relation in standard form.
The inital values are
q[0] = p[1] + p[2] + ... = 1
because of the normalization condition.
And
q[1] = 1 - p[1] = 1 - a p[0]
can be considered as a free parameter in the interval (0,1).
Before we solve (4) we observe that it defines only the elements with an odd index. Therefore we let generally
q[2k] = c[k] > 0, k = 1, 2, ...
with positive c[i].
Performing now the first few steps of the solution to (4) we find that
p[10] = - c[5] - a (1+a) q[1]
But this is a negative quantity, and the contradiction proves the statement.