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The problem is: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$

The first thing I did was use the divergence test which didn't help since the result of the limit was 0.

If I multiply it through, the result is $\sum_{n=1}^{\infty} \frac{1}{n^2+3n}$

I'm wondering if I can consider this as a p-series and simply use the largest power. In this case the power would be 2 which would mean it converges. If this is the correct way to go about this, how do I find where it converges to.

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  • $\begingroup$ Convergence can be dealt with with p-series. Are you wanting to find the sum? $\endgroup$
    – Git Gud
    Commented Dec 2, 2013 at 0:10
  • $\begingroup$ Yes, if it is convergent, I would like to find its sum. $\endgroup$ Commented Dec 2, 2013 at 0:13
  • $\begingroup$ @ConfusingCalc use Partial fractions! $\endgroup$
    – Alec Teal
    Commented Dec 2, 2013 at 0:14

4 Answers 4

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Bound it above! Note $n(n+3)=n^2+3n>n^2$

so $\frac{1}{n(n+3)}<\frac{1}{n^2}$

Each term is clearly > 0 btw.

So! $\sum\frac{1}{n(n+3)}<\sum\frac{1}{n^2}$ which you ought to know (but can trivially show) converges.

Finally a question I can answer here!

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  • $\begingroup$ I'm currently reading up on using partial fractions to get my sum. Are you able to edit it into your answer so I can check my result against yours when I get done with reading about it and trying it? $\endgroup$ Commented Dec 2, 2013 at 0:23
  • $\begingroup$ @ConfusingCalc math.stackexchange.com/a/588808/66223 it's there. With partial fractions you'll get $\frac{1}{n(n+3)}$ the tripple-equal meaning "always equal to, an identity" $\frac{A}{n}+\frac{B}{n+3}$ - find A and B, it turns out N is negative, so you get a "telescoping series" where terms cancel out, almost every term cancels with another. $\endgroup$
    – Alec Teal
    Commented Dec 2, 2013 at 0:25
  • $\begingroup$ I think I incorrectly solved my partial sum. $\frac{1}{n(n+3)} = A(n+3) + B(n)$ How does n turn negative? I wish Hagen would have gone slightly more in depth on how he got what he did. $\endgroup$ Commented Dec 2, 2013 at 1:08
  • $\begingroup$ You should have gotten A=1/3 and B=-1/3, anyway you now need to click this link: lmgtfy.com/?q=telescoping%20sums, it'll help if you write the sum out as the first 5 terms then some ...s and the last 5 @ConfusingCalc $\endgroup$
    – Alec Teal
    Commented Dec 2, 2013 at 1:10
  • $\begingroup$ Thanks for the extra effort. Off to read that link! $\endgroup$ Commented Dec 2, 2013 at 1:18
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Note that $\frac1{n(n+3)}=\frac13\left(\frac1n-\frac1{n+3}\right)$ so this is a telescoping sum $$\sum_{n=1}^m \frac1{n(n+3)}=\frac13\left(1+\frac 12+\frac13-\frac1{m+1}-\frac1{m+2}-\frac1{m+3}\right)\to \frac{11}{18}.$$

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  • $\begingroup$ It took me awhile to get back to you, but after watching a video tutorial on this, I now understand where all this information came from. Thank you! $\endgroup$ Commented Dec 2, 2013 at 18:05
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First, use estimations

$$ n^2 + 3n \geq n^2 \implies \frac{1}{n^2 + 3n } \leq \frac{1}{n^2} $$

Secondly, show that $\sum \frac{1}{n^2}$ converges. In fact, it does. More generally,

$$ \sum \frac{1}{n^p} \; \; \text{converges when} \; \; p > 1 $$

Third, use the comparison theorem: if $a_n \geq b_n $ for all $n$ and $\sum a_n$ converges, then $\sum b_n$ must converge as well (Proof?)

Now, as an application of this theorem, with $a_n = \frac{1}{n^2} $ and $b_n = \frac{1}{n^2 + 3n}$, we notice that your series

$$ \sum \frac{1}{n^2 + 3n} $$

must converge.

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  • $\begingroup$ Thank you for taking the time to answer :) after going through multiple tutorials your answer makes perfect sense! $\endgroup$ Commented Dec 2, 2013 at 18:04
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$$n^2 + 3n > n^2 \implies \frac{1}{n^2 +3n} < \frac{1}{n^2}$$

Use the Comparison Test which states that if $\sum a_n$ and $\sum b_n$ are such that $0 \le a_n \le b_n$, if $\sum b_n$ converges, then $\sum a_n$ converges.

Since $0 < \sum \frac{1}{n^2 +3n} < \sum \frac{1}{n^2}$ and $\sum \frac{1}{n^2}$ converges, then $\sum \frac{1}{n^2 +3n}$ converges.

Edit. Note: $\sum \frac{1}{n^2}$ converges since it is a p-series $$f(x) = \frac{1}{X^p}$$ with $p > 1$ and hence it converges.

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    $\begingroup$ Thank you for showing me this, I didn't realize what I was attempting to do was simply the comparison test. $\endgroup$ Commented Dec 2, 2013 at 18:03
  • $\begingroup$ No problem. My current professor is a stickler for theorems so I always try to provide all details in prep. for his tests. $\endgroup$
    – Zhoe
    Commented Dec 2, 2013 at 18:21

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