Convergence of a series I have to search for convergence of the series 
$\sum_{n=0}^{\infty} \frac{n+1}{n^2+2n+3}$.
I tried to do this by ratio and root test, but unfortunately both are one.
I already found out that it is divergent.
So I tried to show this by comparison test.
I wanted do to this by the estimation that
$\frac{n+1}{n^2+2n+3}\ge \frac{1}{n+3} $, but obviously that didn't help.
Can someone help me, please?
 A: We have $$\frac{n+1}{n^2+2n+3}> \frac{n+1}{n^2+4n+3}=\frac{n+1}{(n+1)(n+3)}=\frac1{n+3}, $$
and $\sum_n \frac1{n+3}$ does not converge (by integral test or limit comparison with $\sum_n\frac1n$). It follows by basic comparison test that $\sum_n \frac{n+1}{n^2+2n+3}$ does not converge.
A: Direct comparison should work here, but you may also use the limit comparison test. Given two series $S_{a}$ and $S_{b},$ if $\lim_{n\to\infty}\frac{S_{b}}{S_{a}} > 0$ and exists, then they are either both diverging or converging. Let us compare
$$S_{a} = \sum_{n = 0}^{\infty}\frac{n + 1}{n^{2} + 2n + 3}$$
with the series
$$S_{b} = \sum_{n = 0}^{\infty}\frac{1}{n}.$$
By the limit comparison test,
$$\lim_{n\to\infty}\frac{S_{b}}{S_{a}}$$
$$= \lim_{n\to\infty}\frac{n^{2} + 2n + 3}{n^{2} + n}$$
$$= 1 > 0.$$
So since $\sum_{n = 0}^{\infty}\frac{1}{n}$ diverges, the original series also diverges.
A: let $f(x)=\frac{x+1}{x^2+2x+3}$. This function is continues, positive and decreasing in $[1,+\infty)$, so we can calculate following integral instead of convergence of the series.
$$\int_{1}^{\infty}\frac{x+1}{x^2+2x+3}dx=\frac{1}{2}\ln(x^2+2x+3)\mathcal{|_{1}^{+\infty}}\to \infty$$ 
A: You can try finding whether
$$
\frac{n+1}{n^2+2n+3}\ge\frac{1}{n+a}
$$
holds for $n$ sufficiently large, where $a>1$. This becomes
$$
n^2+(a+1)n+a\ge n^2+2n+3
$$
so
$$
n\ge\frac{3-a}{a-1}
$$
You see that taking $a\ge3$ is good.
