能否在未知通项公式的情况下判断给定级数的敛散性？

Great question! Let's work through this together—you don't actually need the given explicit formula to figure out if this series converges or diverges. Here's how to approach it:

第一步：观察项的规律 #

First, look at the denominators of the series terms: 2, 6, 10, 14, 18... Notice that each denominator increases by a fixed value (4) every time. That means the denominators form an arithmetic sequence（首项=2，公差=4）. Even if you don't write out the full formula for the denominator, you can recognize it's a linear function of ( n ) (since arithmetic sequences have linear通项).

第二步：用比较判别法（不等式形式） #

You mentioned trying comparison with ( 1/n ) and ( 1/n^2 ) without luck—let's adjust the inequality direction:

• For all ( n \geq 1 ), the ( n )-th denominator can be derived from the arithmetic sequence rule: ( d_n = d_1 + (n-1)\times\text{公差} = 2 + 4(n-1) = 4n-2 ).
• Since ( 4n - 2 \leq 4n ), taking reciprocals reverses the inequality: ( \frac{1}{4n - 2} \geq \frac{1}{4n} ).
• Now, consider the series ( \sum_{n=1}^\infty \frac{1}{4n} = \frac{1}{4}\sum_{n=1}^\infty \frac{1}{n} ). The harmonic series ( \sum \frac{1}{n} ) is famously divergent, so scaling it by ( 1/4 ) doesn't change that—it's still divergent.
• By the comparison test: If every term of a series is greater than or equal to the corresponding term of a divergent series, the original series must also diverge. So our target series is divergent.

第三步：用极限比较判别法（更通用的思路） #

If you prefer a more formal approach that doesn't rely on finding the right inequality, use the limit comparison test:

• Take the harmonic series ( \sum \frac{1}{n} ) (known to diverge).
• Calculate the limit:
  [
  \lim_{n\to\infty} \frac{\frac{1}{4n - 2}}{\frac{1}{n}} = \lim_{n\to\infty} \frac{n}{4n - 2} = \frac{1}{4}
  ]
• This limit is a positive finite number. The limit comparison test tells us that if this is true, the two series have the same convergence behavior. Since ( \sum \frac{1}{n} ) diverges, our series does too.

为什么之前的比较没生效？ #

You probably ran into trouble because you tried comparing directly to ( 1/n^2 ) (which converges, but our terms are larger than ( 1/n^2 ), so that doesn't help) or used the wrong inequality direction with ( 1/n ). Once you recognize the denominators grow linearly, you know the terms behave like ( 1/n ) for large ( n )—and that's the key to picking the right comparison series.

内容的提问来源于stack exchange，提问作者user528378