I can prove the third series diverges, but I don't believe it is known whether or not the fourth, or the more general series, converge. (It is a hard problem, see the remarks at the end)
There is a very good survey paper on integers without large prime factors by Hildebrand and Tenenbaum. They go over the history of results about $\psi(x,y),$ and the Dickmann Debruijn function. For your particular question, we are interested in
$\psi_F(x,y):=\left|\left\{ 1\leq n\leq x:\ P\left(F(n)\right)\leq y\right\} \right|$ where $F$ is any integer valued polynomial. There are few results regarding general polynomials, and some regarding particular cases, fortunately one of which fits exactly this problem. In Hildebrand's paper On integer sets containing strings of consecutive integers, he proves that for $F(x)=\prod_{1\leq j\leq k}(x+j)$, we have $\psi_F(x,x^\alpha)\asymp_{\alpha,k} x$ provided $\alpha>e^{-1/(k-1)}.$ Now, I could not access the paper on the Cambridge journal site, so I have not actually read it, but this result is referred to in the related paper Polynomials values free of large prime factors by Dartyge, Martin and Tenenbaum which is where I first saw it.
Divergence of third series: For $k=2$, the number of $n\leq x$ such that $n(n+1)$ has no prime factors greater then $x^{1/e+\delta}$ is bounded below by a positive constant multiple of $x$. From this it follows that the set $S:=\{n:\ P(n),\ P(n+1)\leq n^{\frac{1}{e}+\delta}\}$ is dense in the integers.** Since $\frac{1}{P(n)P(n+1)}\geq \frac{1}{P(n(n+1))^2},$ by restricting the sum to the set $S$, we are summing $\frac{1}{n^{0.74}}$ over dense subset of integers, and hence the series diverges. (A monotonic positive sequence whose series diverges, still diverges when restricted to any dense subset)
The problem for the fourth series and higher: We could try the same thing for the fourth series, and look at $k=3$ above. However, we can only take $\alpha=\frac{1}{\sqrt{e}}\approx 0.6$, and we have three terms which takes us to $n^{1.8}$, and that series converges. If we could lower the bound on $\alpha$ in Hildebrand's Theorem to $\alpha>\frac{1}{k+\delta}$ instead of $\alpha>e^{-1/(k-1)}$, then the series would diverge for any number of consecutive terms. I think this is likely true, and proving it would be interesting, but it is not that easy.
** We have to be careful when switching from $P(n)\leq x^{1/e+\delta}$ on an interval, to showing the set $S$ is dense, since $S$ is defined with the condition $P(n)\leq n^{1/e+\delta}$. To make the switch rigorously, lets take $\delta/2$, and then look at the interval $(x^{1-\delta/2},x)$. We know there will be at least $cx$ integers $n$ such that $n(n+1)$ has no prime factor greater then $x^{1/e+\delta/2}$. For these $n$, the same condition holds with $P(n(n+1))\leq n^{1/e+\delta}$, so there are at least $cx$ integers $n\in S$. This holds for every interval, so $S$ is dense.