$(a) \sum_{n=2}^{\infty} \frac{3}{n^2+2n}$ 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? 
 A: To compute the sum observe
\begin{align}
\frac{1}{n(n+2)} = \frac{1}{2n}-\frac{1}{2(n+2)}
\end{align}
which means
\begin{align}
\sum^\infty_{n=2}\frac{3}{n^2+2n} =&\ \frac{3}{2}\sum^\infty_{n=2}\left( \frac{1}{n}-\frac{1}{n+2}\right)\\
=&\ \frac{3}{2}\sum^\infty_{n=2}\left(\frac{1}{n}-\frac{1}{n+1}\right)+\frac{3}{2}\sum^\infty_{n=2}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)\\
=&\ \frac{1}{2}+\frac{3}{4}.
\end{align}
A: An alternative approach. The given series is absolutely convergent since the main term behaves like $\frac{1}{n^2}$ and $\sum_{n\geq 1}\frac{1}{n^2}$ is convergent by the $p$-test. We have
$$ \frac{3}{n(n+3)}= \int_{0}^{1}\left(x^{n-1}-x^{n+2}\right)\,dx \tag{1}$$
hence the whole series equals
$$ \int_{0}^{1}\sum_{n\geq 2}\left(x^{n-1}-x^{n+2}\right)\,dx = \int_{0}^{1}(x+x^2+x^3)\,dx = H_4-1=\color{red}{\frac{13}{12}}.\tag{2}$$
A: An often used technique in this situation is called partial fraction decomposition; set:
$$\frac{3}{n(n+2)} = \frac{A}{n} + \frac{B}{n+2}$$
and solve for $A$ and $B$. You will find $A = 3/2$ and $B = -3/2$, which means:
$$\frac{3}{n(n+2)} = \frac{3}{2n} - \frac{3}{2(n+2)}$$
The series now "telescopes" as others have just posted. If you have never seen this method before, here is how you actually solve for $A$ and $B$:
Multiplying the first equation by $n(n+2)$ on both sides yields:
$$3 = A(n+2) + Bn \implies 0n + 3 = (A+B)n + 2A$$
The $0n$ is included for emphasis, for this means $A = -B$. Now, all there is to do is observe that you have $2A = 3$, i.e., $A = 3/2$, which yields $B = -3/2$ as desired.

Also, for what it's worth: A different way to show that the series converges is to observe that each term is less than $3/n^2$, which means the series is bounded above by $3 \sum_{n>1}1/n^2$, which is just three multiplied by a convergent series (perhaps what you call a $p$-series). Since the series is monotonically increasing and bounded above, it must converge.
A: 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.
