关于收敛级数定义与调和级数发散性的疑问
Hey there! Let's work through this confusion together—this is a super common mix-up when first learning series, so don't feel bad about it.
First off, let's clarify a critical definition you might have mixed up:
The statement you quoted ("the series is said to be convergent when n approaches infinite an=L, where L is a constant") is actually the definition for a convergent sequence, not a convergent series.
For a series $\sum_{n=1}^\infty a_n$ to be convergent, we need its partial sum sequence to have a finite limit. The partial sum $S_n = a_1 + a_2 + ... + a_n$ must approach some constant $L$ as $n \to \infty$. That's the real definition of a convergent series.
Now, here's the key point you're missing: The fact that $a_n \to 0$ as $n \to \infty$ is a necessary condition for the series to converge (meaning if the series converges, $a_n$ must go to 0), but it's not a sufficient condition (meaning just because $a_n$ goes to 0 doesn't guarantee the series converges).
Let's break down your examples:
- The sequence $a_n = \frac{n}{n+1}$ does converge to 1, but that's just a sequence, not a series. If you were looking at the series $\sum_{n=1}^\infty \frac{n}{n+1}$, it would actually diverge because the terms don't even approach 0—they approach 1, which violates the necessary condition for convergence.
- For the harmonic series (P=1 P-series) $\sum_{n=1}^\infty \frac{1}{n}$, the terms $a_n = \frac{1}{n}$ do approach 0, but the partial sums grow without bound. A simple way to see this is by grouping the terms:
- $1 + (\frac{1}{2}) + (\frac{1}{3}+\frac{1}{4}) + (\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}) + ...$
- Each group is greater than $\frac{1}{2}$: the first group is 1, the second is $\frac{1}{2}$, the third is $\frac{1}{3}+\frac{1}{4} > \frac{1}{4}+\frac{1}{4} = \frac{1}{2}$, the fourth group has four terms each bigger than $\frac{1}{8}$, so their sum is > $4*\frac{1}{8} = \frac{1}{2}$, and so on. Adding infinitely many terms each at least $\frac{1}{2}$ gives an infinite total, so the partial sums go to infinity.
So to wrap up: Your confusion comes from mixing up the definitions of convergent sequences and convergent series. The harmonic series is a perfect example of why "terms approaching 0" isn't enough for a series to converge—it's a necessary check, but you need more tools (like the integral test, comparison test, ratio test, etc.) to prove convergence.
备注：内容来源于stack exchange，提问作者Ili a
