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So I have this series:


I noticed:


So can I say that it's convergent already or should I use a critera on the second fraction in the sum?

Also for sum I could use partial fractions or am I mistaken?

Any help would be appreciated.

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Note that $n^2+6n+9=(n+3)^2$. Also, Wolfram Mathematica gives $\frac{115}{36}$ for the limit of your sum. – vrugtehagel Feb 15 at 14:03
also i can take 3 out of the numerator – MathIsTheWayOfLife Feb 15 at 14:05
Unless you are looking to find the exact sum, you should be able to do a direct comparison test to something like $\sum\frac{1}{n^2}$. If finding how to correctly use the inequalities in this case is difficult, then a ratio test might be easier. – JMoravitz Feb 15 at 14:08
thank you , i appreciate it – MathIsTheWayOfLife Feb 15 at 14:11
And... $$\sum_{n=1}^{\infty}\frac{1}{n^2}\cdot\frac{3n^2+15n+9}{n^2+6n+9} = \sum_{n=1}^{\infty}\frac{3}{n^2}\cdot\frac{n^2+5n+3}{n^2+6n+9} = 3\sum_{n=1}^{\infty}\frac{1}{n^2}\left(1 - \frac{n+6}{n^2+6n+9}\right)$$ ...which may help you, or may be not. – CiaPan Feb 15 at 14:59
up vote 10 down vote accepted

You can conclude that the series converges by the comparison test. We can also calculate the sum. Note that $$\frac{3n^{2}+15n+9}{n^{4}+6n^{3}+9n^{2}}=\frac{1}{n^{2}}+\frac{1}{n}-\frac{1}{n+3}-\frac{1}{\left(n+3\right)^{2}} $$ hence $$\sum_{n\geq1}\frac{3n^{2}+15n+9}{n^{4}+6n^{3}+9n^{2}}=\sum_{n\geq1}\left(\frac{1}{n^{2}}+\frac{1}{n}-\frac{1}{n+3}-\frac{1}{\left(n+3\right)^{2}}\right)$$ $$=2+\frac{1}{2}+\frac{1}{4}+\frac{1}{3}+\frac{1}{9} =\frac{115}{36} $$ because the series telescopes.

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\begin{align} a_n=\frac{3n^2+15n+9}{n^4+6n^3+9n^2} &\le \frac{3n^2+15n^2+9n^2}{n^2(n+3)^2} \\ &= \frac{27}{(n+3)^2}\\ &\le \frac{27}{n^2} = b_n \end{align}

It is well know that

$$\sum_{n=1}^\infty \frac{1}{n^2}$$

converges. Since $0\le a_n\le b_n$ and $\sum b_n$ converges, we may conclude that $\sum a_n$ converges.

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