含质数判定函数的交替无穷级数：敛散性证明及闭式解问询

Alright, let's tackle this problem step by step—first proving whether the series converges, then addressing the closed-form question.

1. Proving Convergence/Divergence #

First, let's rewrite the series to make its behavior easier to analyze. Remember that $f(n)$ is 1 when $n$ is prime, 0 otherwise, so $(-1)^{f(n)}$ is $-1$ for primes and $+1$ for non-primes. We can split the series into two separate sums:
$$
\sum_{n=2}^{\infty} \frac{(-1)^{f(n)}}{n2} = \sum_{\substack{n=2 \ n \text{ prime}}}^{\infty} \frac{-1}{n^2} + \sum_{\substack{n=2 \ n \text{ not prime}}}^{\infty} \frac{1}{n^2}
$$

Now, let's combine these into a single expression relative to the Riemann zeta function $\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}$. The sum of $1/n^2$ for all $n \geq 2$ is $\zeta(2) - 1$. Notice that our original series is equal to this full sum minus twice the sum of $1/n^2$ over primes:
$$
\sum_{n=2}^{\infty} \frac{(-1)^{f(n)}}{n2} = \left(\zeta(2) - 1\right) - 2\sum_{\substack{n=2 \ n \text{ prime}}}^{\infty} \frac{1}{n^2}
$$

Why twice? Because we're taking the non-prime terms (which are already $+1/n^2$) and adjusting the prime terms from $+1/n^2$ to $-1/n^2$—that's a difference of $-2/n^2$ per prime.

Now, we know two key facts:

• $\zeta(2) = \frac{\pi^2}{6}$, which is a finite, convergent value.
• The sum of $1/p^2$ over all primes $p$ is convergent. This is because it's a subset of the convergent series $\sum_{n=1}^\infty 1/n^2$, so by the comparison test, it must converge.

Since we're taking the difference of two convergent series, the original series is convergent.

2. Does a Closed-Form Solution Exist? #

Short answer: There's no known elementary closed-form (i.e., using basic functions like polynomials, trigonometric functions, logarithms, or simple zeta function combinations). However, we can express it using special functions.

Define the prime zeta function $P(s) = \sum_{p \text{ prime}} \frac{1}{p^s}$. Using this, our series becomes:
$$
\sum_{n=2}^{\infty} \frac{(-1)^{f(n)}}{n2} = \left(\zeta(2) - 1\right) - 2P(2)
$$

Substituting the known value of $\zeta(2)$:
$$
= \frac{\pi^2}{6} - 1 - 2P(2)
$$

While we can compute the numerical value of $P(2)$ (it's approximately 0.4522), there's no way to write $P(2)$ using elementary functions. So the series can be expressed in terms of special functions, but it doesn't have an elementary closed-form solution.

内容的提问来源于stack exchange，提问作者Razvan