<s>[INST] $(a) \sum_{n=2}^{\infty} \frac{3}{n^2+2n}$ Determine whether it converges and find sum Determine whether the following are convergent or divergent. If it is
convergent find its sum. Make sure to fully justify all of your work.
$$(a) \sum_{n=2}^\infty \frac{3}{n^2+3n}$$
Solution (My attempt):
Let $f(n) = a_n$
On $[2,\infty), f(x) = \frac 3 {x^2+3x} > 0$
$$f'(x) = 3\left(\frac{-(2x+3)}{(x^2+3x)^2}\right) < 0, \forall x \in [2,\infty)$$
Therefore the hypothesis of the integral test is met.
Consider
$$\lim_{A\to\infty} \int_2^A \frac 3 {(x)(x+3)} \, dx$$
$$\lim_{A\to\infty} \int_{2}^{A} \left( \frac{1}{x} - \frac{1}{x+3} \right)dx \text{ By pfd}$$
$$\lim_{A\to\infty} \left( \ln \right(\frac A {A+3}\left) - \ln \right(\frac 2 5 \left)\right) $$
$$= \ln 1 - \ln \left(\frac{2}{5}\right)$$
Therefore by Integral test the original series converges.
I was never thought how to find the sum of this type of series. The only way I know how to find the sum of a series is if its geometric. Does anyone know how too find the sum? 
 [/INST] Consider the partial sum $$S_p=\sum_{n=2}^p\frac{3}{n^2+3n}=\sum_{n=2}^p\frac{1}{n}-\sum_{n=2}^p\frac{1}{n+3}$$ $$S_p=\left(\frac 12+\frac 13+\frac 14+\frac 15+\cdots+\frac 1p\right)-\left(\frac 15+\frac 16+\frac 17+\frac 18+\cdots+\frac 1{p+3}\right)$$ What is left is $$S_p=\frac{13}{12}-\frac{1}{p+1}-\frac{1}{p+2}-\frac{1}{p+3}=\frac{13}{12}-\frac{3 p^2+12 p+11}{p^3+6 p^2+11 p+6}$$ Now, for large $p$, perform the long division (or Taylor series) to get $$S_p=\frac{13}{12}-\frac{3}{p}+\frac{6}{p^2}-\frac{14}{p^3}+O\left(\frac{1}{p^4}\right)$$ For $p=10$, $S_{10}=\frac{119}{143}\approx 0.8322$ while the expansion gives $\frac{311}{375}\approx 0.8293$.
You also could use the harmonic numbers and arrive to $$S_p=\frac{13}{12}+H_p-H_{p+3}$$ and using the asymptotics $$H_n=\gamma +\log \left({n}\right)+\frac{1}{2 n}-\frac{1}{12
   n^2}+O\left(\frac{1}{p^4}\right)$$ and arrive to the same result.
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