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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?

SR23
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    Why didn't that help? It should settle the matter. – Daniel Fischer Jun 01 '16 at 20:55
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    Note that if $n\gt 0$ then $\frac{n+1}{n^2+2n+3}\gt \frac{n}{n^2+2n^2+3n^2}=\frac{1}{6n}$. Or else do a limit comparison with $\sum_1^\infty \frac{1}{n}$. – André Nicolas Jun 01 '16 at 20:57
  • I am just aware that $\frac{1}{n}$ is divergent, but this means not that $\frac{1}{n+3}$ is divergent, or am I wrong? – SR23 Jun 01 '16 at 20:58
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    @SR23 You are wrong, if $\sum\frac1n$ diverges, then $\sum\frac1{n+3}$ also diverges. Anything of the form $\sum\frac1{an+b}$ diverges, and more generally, $\sum\frac{an+b}{pn^2+qn+r}$ diverges. – Simply Beautiful Art Jun 01 '16 at 21:03
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    The series for $\frac{1}{n+3}$ is the same series as for $\frac{1}{n}$, you're just starting a little later ($\sum_{n=0}^\infty\frac{1}{n+3}=\sum_{n=3}^\infty\frac{1}{n}$). – Michael Burr Jun 01 '16 at 21:09

4 Answers4

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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.

Math1000
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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.

K. Jiang
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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. $$\int_{1}^{\infty}\frac{x+1}{x^2+2x+3}dx=\frac{1}{2}\ln(x^2+2x+3)\mathcal{|_{1}^{+\infty}}\to \infty$$

Behrouz Maleki
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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.

egreg
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