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Determine if the series converges or diverges. If it converges what does it converge to?

${\sum_{n=1}^\infty \frac{4}{n(n+3)}}$

Here's what I have been able to figure out

By the limit comparison test ${\sum\frac{1}{x^2}}$

${\lim_{x\to\infty}\frac{4n^2}{n^2 + 3n} = 4\Rightarrow}$ Convergence

But I can't figure out how to tell what it converges to, it's not an alternating series, or a geometric series that I can tell. And we haven't yet covered taylor series. It might be a power series, but I don't see any x value to tell what it converges to.

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\begin{align*} &\dfrac{4}{3}\displaystyle\sum_{n}\left(\dfrac{1}{n}-\dfrac{1}{n+3}\right)\\ &=\dfrac{4}{3}\displaystyle\sum_{n}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)+\dfrac{4}{3}\displaystyle\sum_{n}\left(\dfrac{1}{n+1}-\dfrac{1}{n+2}\right)+\dfrac{4}{3}\displaystyle\sum_{n}\left(\dfrac{1}{n+2}-\dfrac{1}{n+3}\right) \end{align*}

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