calculus - Determining if the series $sum_{n=1}^infty {nover {n^2+1}}$ is convergent or divergent and justifying the answer.

I'm trying to solve the problem above, but I have gotten stuck. I can't use p-series test easily to check it. Maybe the integral test? Here's a picture of some rules my book gives me, just in case anyone needs to know what I have learned already:

Going by those rules, step 3 seems to be my only option. Going by that, $$a_n={n\over{n^2+1}}\;\;and\;\;b_n={n\over{n^2}}$$ Since $P=2$ that means $P>1$ and it converges. So, by the comparison test, ${n\over{n^2+1}}$ converges. Does that seem correct? I feel like that is way to easy, but I don't know.

Thank you.

Answer

You're right that step 3 is the way to go. You'd like to compare to
$$
b_n = \frac{n}{n^2} = \frac{1}{n}
$$
Then $a_n < b_n$, and $\sum b_n$ is a $p$-series. But $p=1$, so the series $\sum b_n$ diverges. This means our regular comparison test won't work. It's not useful to say a series is less than a divergent one—it's like saying the sum of the series is less than or equal to $\infty$.

There are two options:

1. Compare $a_n$ to $b_n = \frac{1}{2n}$. Hopefully for all $n$, or at least for all $n$ past some number $n_n$, $a_n > b_n$. And $\sum b_n$ still diverges since all we did is scale by a constant. So the comparison test gives you what you want: $\sum a_n$ diverges.




2. Use the Limit Comparison Test (this may be something you have covered, or it may not be). Let $b_n = \frac{1}{n}$ as before. Then since
   $$
   \lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty} \frac{n^2}{n^2+1} = 1
   $$
   the series $\sum a_n$ and $\sum b_n$ either both converge or both diverge. Since $\sum b_n$ diverges, we know $\sum a_n$ diverges too.

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A word from the math grammar police: Please be sure to distinguish between a series and the sequence of its terms. We sometimes say things like “$b_n$ diverges” when we mean the series $\sum b_n$ diverges. The sequence $b_n$ actually converges (to zero). The $\sum$ operator is an important part of the expression. Skipping over it is like confusing a function and a definite integral. It's at best sloppy, and at worst confusing.