Determining if the series $\sum_{n=1}^\infty {n\over {n^2+1}}$ 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.
 A: $b_n$ is the harmonic series (cancel the $n$'s), or a $p$-series with $p=1$, which diverges.  Unfortunately $a_n$ is smaller, so you need to do the limit comparison test (not just the ordinary comparison test).  It turns out that $a_n$ and $b_n$ do match, so $a_n$ diverges too.
A: 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:


*

*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.

*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.

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.
A: Note that 
$$
\frac{n}{n^2 + 1} \ge \frac{n}{n^2 + n} \ge \frac{n}{n^2 + n^2} = \frac{1}{2n},
$$
hence, 
$$
\sum_{n=1}^{\infty}\frac{n}{n^2 + 1}   \ge \frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{n}  . 
$$
A: use that $$\frac{n}{n^2+1}\geq \frac{1}{2n}$$
