Let $p_n$ be the $n$-th prime number (starting with $p_0=0 , p_1=2, ...$). I computed the zeros of the polynomial $$ f_n(x) = p_nx^n + p_{n-1}x^{n-1} + ... + p_1x + p_0 $$ for different $n$. This polynomial has one real root $\tilde{x}_n\neq 0$ for $n$ even. Here are some example values: $$ f_2(x) =0 ~,~ \tilde{x}_2=-0.666666... \\ f_{6}(x) =0 ~,~ \tilde{x}_6=-0.750523... \\ f_{12}(x) =0 ~,~ \tilde{x}_{12}=-0.802329...\\ f_{14}(x) =0 ~,~ \tilde{x}_{14}=-0.823575... \\ (f_{46}(x) =0 ~,~ \tilde{x}_{46}=-0.925623...) \\ (f_{96}(x) =0 ~,~ \tilde{x}_{96}=-0.963122...) \\ $$ The values appear to approach $-1$. This raises the question if $$ \lim_{m\rightarrow\infty} \sum^{2m}_{n=0} (-1)^n p_n = 0. $$ Is that true? How to prove it?
EDIT TO CLARIFY MY QUESTION: Take a look at the function $$f_n(x) = \sum_{k=0}^n p_k x^k,$$ i.e. a polynomial of degree $k$ with prime numbers as coefficients. Task: Calculate the real roots of the polynomial for different $k$. For even $n$ we do not have any real roots. For odd $n$ we do have one real root $x_0(n)$, i.e. for any $m$ we have the root $x_0(2m+1)$. The value of this root $x_0(n)$ is given above and in the table below for increasing $n$. As you can see, it seems that the values are approaching $-1$ for increasing $m$, i.e. we have $$\lim\limits_{m\rightarrow \infty}x_0(2m+1) = -1.$$ That is, for $m\rightarrow \infty$ we would have $f_{2m+1}(-1) = 0$ or $$ \lim\limits_{m\rightarrow \infty}f_{2m+1}(-1) = 0 $$ Now my question is the following: Does the root of a prime polynomial with odd degree tends to be $-1$ for increasing degree?
Ok, since this question is on hold and I can't write a comment or answer properly to this question, here's my answer to your problem:
It's not true! Look at this the following way: $$f_{2m}(-1)=\sum^{2m}_{n=0} (-1)^n p_n=\sum^{2m}_{n=1} (-1)^n p_n=\sum^{m}_{n=1} p_{2n}-p_{2n-1}\geq m$$ since $p_0=0$ and $p_{2n}-p_{2n-1}\geq 1$ for all $n$. So obviously the alternating sum over $2m$ primes diverges! The pattern you noticed would imply $$\lim_{m\rightarrow\infty} \tilde{x}_{2m}=-1$$ I don't checked if it's true, but I think it is.
Another fact is obviously $f_{2m}(\tilde{x}_{2m})=0$ since $\tilde{x}_{2m}$ is a root of exactly this polynomial. But even if the above limit of the roots is correct, here is where it breaks: $$0=\lim_{m\rightarrow\infty} f_{2m}(\tilde{x}_{2m})$$ $$\neq$$ $$ \lim_{m\rightarrow\infty} \lim_{k\rightarrow\infty}f_{2m}(\tilde{x}_{2k})=\lim_{m\rightarrow\infty} f_{2m}(\lim_{k\rightarrow\infty} \tilde{x}_{2k})=\lim_{m\rightarrow\infty} f_{2m}(-1)=\lim_{m\rightarrow\infty} \sum^{2m}_{n=0} (-1)^n p_n=\infty$$ The mistake made is in splitting one limit in two independent limits.