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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.

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    $\begingroup$ 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. $\endgroup$ – G Tony Jacobs Sep 1 '17 at 21:59
  • $\begingroup$ I'll do the integral test, but it is still convergent, no? $\endgroup$ – Shea Sep 1 '17 at 22:00
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    $\begingroup$ Compare to $\sum\limits_{n=1}^\infty \frac{1}{n^2}$. Which is bigger. Do you know anything about the sum of reciprocal squares? $\endgroup$ – JMoravitz Sep 1 '17 at 22:02
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    $\begingroup$ Hint : The partial fraction decomposition allows to derive a telescope sum. $\endgroup$ – Peter Sep 1 '17 at 22:03
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    $\begingroup$ 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. $\endgroup$ – JMoravitz Sep 1 '17 at 22:05
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\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}

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  • $\begingroup$ 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.) $\endgroup$ – Daniel Robert-Nicoud Sep 1 '17 at 23:18
  • $\begingroup$ @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. $\endgroup$ – José Carlos Santos Sep 1 '17 at 23:31
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$$S=\sum\limits_{n=1}^\infty\frac{1}{n^2+5n+6}=\sum\limits_{n=1}^\infty\frac{1}{(n+2)(n+3)}=\sum\limits_{n=1}^\infty \left(\frac{1}{n+2}-\frac{1}{n+3}\right)$$

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

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

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$\lim_{n \to \infty} a_n = 0$ 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 $\frac{1}{3}$.

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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 $\frac{1}{3}$.

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

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

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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

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

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