# Determine whether the series $\sum\limits_{n=1}^\infty\frac{1}{n^2+5n+6}$ is convergent or divergent. If it is convergent, find its sum.

Determine whether the follwing series is convergent or divergent. If it is convergent, find the sum. $$\sum\limits_{n=1}^\infty\frac{1}{n^2+5n+6}$$

Here's my work so far:

$\lim_\limits{n\to\infty}\frac{1}{n^2+5n+6} = 0$

$\therefore\;$ the series is convergent.

I don't think it's a geometric series since there is no common factor between the consecutive terms. Because it isn't a geometric series, I'm at a loss as to what formula to use.

• Showing that the terms go to zero does not establish that the series is convergent. The terms of the harmonic series go to zero, and it diverges. Sep 1, 2017 at 21:59
• I'll do the integral test, but it is still convergent, no?
– Shea
Sep 1, 2017 at 22:00
• Compare to $\sum\limits_{n=1}^\infty \frac{1}{n^2}$. Which is bigger. Do you know anything about the sum of reciprocal squares? Sep 1, 2017 at 22:02
• Hint : The partial fraction decomposition allows to derive a telescope sum. Sep 1, 2017 at 22:03
• But yes, as GTonyJacobs already alluded to $\sum\limits_{n=1}^\infty f(n)$ converging implies that $\lim\limits_{n\to\infty}f(n)=0$ however the converse is not true. That is to say, knowing $\lim\limits_{n\to\infty}f(n)=0$ we do not know anything about whether or not $\sum\limits_{n=1}^\infty f(n)$ converges or diverges. The contrapositive however can give us useful information, that is to say if we know $\lim\limits_{n\to\infty}f(n)\neq 0$, either because it converges to something else or doesn't converge at all, then we can know that $\sum\limits_{n=1}^\infty f(n)$ doesn't converge. Sep 1, 2017 at 22:05

\begin{align}\sum_{n=1}^\infty\frac1{n^2+5n+6}&=\sum_{n=1}^\infty\frac1{(n+2)(n+3)}\\&=\sum_{n=1}^\infty\left(\frac1{n+2}-\frac1{n+3}\right)\\&=\left(\frac13-\frac14\right)+\left(\frac14-\frac15\right)+\left(\frac15-\frac16\right)+\cdots\\&=\frac13-\lim_{n\to\infty}\frac1{n+2}\\&=\frac13.\end{align}

• In order to do such manipulations, you first need to prove that the series converge! (Well, it does work in this case of course, I just wanted to stress the fact.) Sep 1, 2017 at 23:18
• @DanielRobert-Nicoud I've just added one line. Anyway, the only thing missing was that $\sum_{n=1}^\infty(a_n-a_{n+1})$ converges if and only if the limit $\lim_{n\to\infty}a_n$ exists and it is real. Sep 1, 2017 at 23:31

$$S=\sum\limits_{n=1}^\infty\frac{1}{n^2+5n+6}=\sum\limits_{n=1}^\infty\frac{1}{(n+2)(n+3)}=\sum\limits_{n=1}^\infty \left(\frac{1}{n+2}-\frac{1}{n+3}\right)$$

$S=\mathop {\lim }\limits_{n \to \infty } S_n$

where $S_n$ are the partial sums

$S_1=a_1=\dfrac{1}{3}-\dfrac{1}{4}$

$S_2=a_1+a_2=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}=\dfrac{1}{3}-\dfrac{1}{5}$

$S_3=a_1+a_2+a_3=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{1}{3}-\dfrac{1}{6}$

$\cdots\cdots$

$S_n=\dfrac{1}{3}-\dfrac{1}{n+3}$

$S=\mathop {\lim }\limits_{n \to \infty }\left(\dfrac{1}{3}-\dfrac{1}{n+3}\right)=\dfrac{1}{3}$

Hope it helps

$\lim_{n \to \infty} a_n = 0$ isn't sufficient to prove the series $\sum_{n=1}^{\infty} a_n$ is convergent, take $a_n = \frac{1}{n}$ for instance.

Note that $$\frac1{n^2 + 5n+6} = \frac{1}{(n+2)(n+3)} = \frac1{n+2} - \frac1{n+3}$$ So, \begin{align}\sum_{n=1}^{N}\frac1{n^2 + 5n+6} &= \sum_{n=1}^{N}\left(\frac1{n+2} - \frac1{n+3}\right)\\ &= \frac{1}{3}-\frac{1}{N+3}\end{align}

From here, you can see that as $N \to \infty$, the series converges, and that it converges to $\frac{1}{3}$.

It is convergent & can be summed telescopically after doing some partial fractions \begin{eqnarray*} \frac{1}{n^2+5n+6} = \frac{1}{n+2}- \frac{1}{n+3} \end{eqnarray*} So the sum is $\frac{1}{3}$.

$$\frac{1}{n² + 5n + 6} < \frac{1}{n²}$$

And $\sum \frac{1}{n²}$ is well known as convergent, so $\sum \frac{1}{n² + 5n + 6}$ must be as well by comparison.

hint

$$\int_n^{n+1}\frac{dx}{x^2+5x+6}<\frac {1}{n^2+5n+6}<\int_{n-1}^n\frac {dx}{x^2+5x+6}$$

and

$$\int \frac {dx}{x^2+5x+6}=\ln (\frac {x+2}{x+3} )$$