Prove that $\sum_{n=1}^\infty\frac{n}{n^2+1}$ is divergent

I'm trying to show that the series $\sum_{n=1}^\infty\frac{n}{n^2+1}$ is divergent.

I'm trying to use a comparison test to do so. To do so, I first can see that $\sum_{n=1}^\infty\frac{n}{n^2+1} \geq \sum_{n=1}^\infty\frac{n}{n^2+n} = \sum_{n=1}^\infty\frac{1}{n+1}$ And I can use a change of variable $k = n+1$, so show $\sum_{n=1}^\infty\frac{1}{n+1} = \sum_{k=2}^\infty\frac{1}{k} = \infty$. Then since $\sum_{n=1}^\infty\frac{n}{n^2+1} \geq \sum_{k=2}^\infty\frac{1}{k}$ it must be that $\sum_{k=2}^\infty\frac{1}{k}$ also diverges to infinity. Is this a valid way to show it?

Solution 1:

Well, I will improve your solution by saying that $$\frac{n}{n^2+\color\red1}\geq\frac{n}{n^2+\color\red{n^2}}=\frac{1}{2}\cdot\frac{1}{n}\qquad\forall n\in\Bbb N.$$ We know that series (called as harmomic series) $$\sum_{n=1}^\infty\frac{1}{n}$$ is divergent and so multiplying it by $\frac{1}{2}$, it follows that the series $$\sum_{n=1}^\infty\bigg(\frac{1}{2}\cdot\frac{1}{n}\bigg)$$ is also divergent and hence, it follows from Comparison Test that the series $$\sum_{n=1}^\infty\frac{n}{n^2+1}$$ is divergent.