调和级数变体的收敛性探讨：素数下标素数倒数级数收敛原因

Great question! Let's unpack this by leaning on the Prime Number Theorem (PNT) and basic convergence tests for positive series—key tools here. First, let's clarify notation: $p_n$ denotes the $n$-th prime number (so $p_1=2$, $p_2=3$, $p_3=5$, etc.).

1. Why $\sum\frac{1}{p_n}$ diverges #

The Prime Number Theorem tells us that for large $n$, the $n$-th prime grows like:
$$p_n \sim n\log n$$
This means that as $n$ gets very big, $p_n$ is approximately equal to $n$ times the natural log of $n$. Taking reciprocals, we get:
$$\frac{1}{p_n} \sim \frac{1}{n\log n}$$
Now, recall that the series $\sum\frac{1}{n\log n}$ diverges—we can prove this with the integral test: the integral of $\frac{1}{x\log x}$ from 2 to infinity is $\log(\log x)$, which tends to infinity as $x\to\infty$. Since $\frac{1}{p_n}$ behaves like the terms of a divergent series for large $n$, the comparison test confirms $\sum\frac{1}{p_n}$ must also diverge.

2. Why $\sum\frac{1}{p_{p_n}}$ converges #

Now let's look at $p_{p_n}$—this is the $p_n$-th prime. Using the Prime Number Theorem again, we can approximate its growth rate. Since $p_n \sim n\log n$, substitute this into the PNT formula for the $k$-th prime (where $k=p_n$):
$$p_{p_n} \sim p_n \cdot \log(p_n)$$
Next, substitute $p_n \sim n\log n$ into the right-hand side:
$$p_{p_n} \sim (n\log n) \cdot \log(n\log n)$$
Simplify $\log(n\log n)$: this equals $\log n + \log(\log n)$, and for large $n$, $\log n$ dominates the much smaller $\log(\log n)$. So we can approximate:
$$\log(n\log n) \sim \log n$$
Putting it all together:
$$p_{p_n} \sim n\log n \cdot \log n = n(\log n)^2$$
Taking reciprocals:
$$\frac{1}{p_{p_n}} \sim \frac{1}{n(\log n)^2}$$
Now, the series $\sum\frac{1}{n(\log n)^2}$ converges—again using the integral test: let $u = \log x$, so $du = \frac{1}{x}dx$. The integral from 2 to infinity becomes:
$$\int_{\log 2}^\infty \frac{1}{u^2}du = \left[ -\frac{1}{u} \right]{\log 2}^\infty = \frac{1}{\log 2}$$
This is a finite value, so the integral converges, meaning the series $\sum\frac{1}{n(\log n)^2}$ converges. By the limit comparison test, since $\frac{1}{p{p_n}}$ behaves like the terms of this convergent series for large $n$, $\sum\frac{1}{p_{p_n}}$ must also converge.

The key takeaway here is that primes grow faster than linear, but the sequence of primes indexed by primes grows much faster—fast enough that their reciprocals form a convergent series.

内容的提问来源于stack exchange，提问作者dahaka5