Prove that the non-trivial root of $\sum_{k=1}^{2n} p_kx^k=0$ tends to $-1$ I looked at
$$
\sum_{k=1}^{2n} p_kx^k=0,
$$
where $p_k$ is the $k$th prime. I found that, next to the trivial root $x_0=0$, there is only one more root $x_n$ that tends towards $-1$, when $n$ increases. My question is:

Can it be proven that $\lim \limits_{n\to \infty} x_n =-1$?

 A: It's true, but it seems to be hard.  I give a sketch of a proof below.   Let $z=-x$, so that our polynomial looks like  $$f(z)=\sum_{k=1}^{n}\left(p_{2k}z-p_{2k-1}\right)z^{k}$$ where $z$  is positive. Our goal is to show that there is a real root that approaches $1$, and to do this we use the intermediate value theorem. Taking $z=1$, since $p_{2k}>p_{2k-1}$, we see that the above sum is positive, that is $f(1)>0$. Now, I claim that for a suitably chosen constant $C>0$, we have that $$f\left(1-\frac{C}{n}\right)<0.$$ If one can prove this, then by the intermediate value theorem, there exists a root of the original polynomial which approaches $-1$. Here is a quick sketch about how the proof should go: On average, we have that $$p_{2k}-p_{2k-1}\sim\log(2k),$$ and so, on average, $$\left(1-\frac{1}{2k}\right)p_{2k}-p_{2k-1}=o\left(\log(2k)\right),$$ which is quite small. This means that for $C>\frac{1}{2},$ on average, $$\left(1-\frac{C}{k}\right)p_{2k}-p_{2k-1}\ll-\log(2k).$$ Moving from an average statement to the whole can be tricky because we are dealing with the primes and because we have a decreasing weight function to handle.  On top of this, we must choose $z=1-\frac{C}{n},$ so the above will hold only for a portion of the $k$, and for small $k$ the difference may be positive on average.   We choose $c>0$  depending on $C$  so that we have the correct inequality for $k>cn$. Then we will split the sum and look at $$\sum_{k>cn}^{n}\left(p_{2k}\left(1-\frac{C}{n}\right)-p_{2k-1}\right)\left(1-\frac{C}{n}\right)^{k}+\sum_{k\leq cn}^{n}\left(p_{2k}\left(1-\frac{C}{n}\right)-p_{2k-1}\right)\left(1-\frac{C}{n}\right)^{k}.$$ The coefficients of the first sum are negative on average, and the second sum is a quantity that we hope to bound outright. To deal with the fact that we only have the coefficients on average, and that we have a weight $$\left(1-\frac{C}{n}\right)^{k},$$ we note that the weight is extremely well behaved on short intervals, so we may appeal to results regarding prime gaps on short intervals. By partitioning into these short intervals, dealing with the average there, and the moving back to the entire sum, we can prove that the first sum above is large and negative. The next goal is to show that the second sum is not too large, and so the entire quantity is negative. 
Edit:  It seems that $C$ must also depend on $n$.
