<s>[INST] Series using partial sums... Determine if the following series converge by studying the partial sums. If it converges, compute its value.
$\sum_{n=1}^{\infty} \frac{2n+3}{(n+1)(n+2)}$.
We use partial fractions so
\begin{equation*}
\begin{split}
\frac{2n+3}{(n+1)(n+2)} &= \frac{A}{n+1}+\frac{B}{n+2} \\
\Longleftrightarrow2n+3 &= A(n+2)+B(n+1) \\
&= An+2A+Bn+B \\
&= (A+B)n+2A+B.
\end{split}
\end{equation*}
Equating terms tells us that $A+B = 2$ and $2A+B = 3$. We can find $A,B$ and $C$ by solving
\begin{equation*}
\begin{bmatrix}
1 & 1 \\
2 & 1
\end{bmatrix}\begin{bmatrix}
A \\
B
\end{bmatrix} = \begin{bmatrix}
2 \\
3
\end{bmatrix}.
\end{equation*}
Doing so yields $A = B = 1$. This means the partial sum is
\begin{equation*}
S_k = \sum_{n=1}^{k} \frac{2n+3}{(n+1)(n+2)} = \sum_{n=1}^{k} \left(\frac{1}{n+1}+\frac{1}{n+2}\right) = \left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+\ldots+\left(\frac{1}{k+1}+\frac{1}{k+2}\right).
\end{equation*}
Is this correct?
 [/INST] Your method is correct,  I just put a simpler approach if you are interested.  We see that $$\frac{2n+3}{(n+1)(n+2)}>\frac{2n}{n^2+3n+2}>\frac{2n} {6n^2}=\frac{1}{3n} $$
But $\sum{\frac{1}{n}}$  diverges..... 
</s>