Compute
\[\sum_{n = 1}^\infty \frac{2n + 1}{n(n + 1)(n + 2)}.\]
First, we decompose $\frac{2n + 1}{n(n + 1)(n + 2)}$ into partial fractions.  Let
\[\frac{2n + 1}{n(n + 1)(n + 2)} = \frac{A}{n} + \frac{B}{n + 1} + \frac{C}{n + 2}.\]Then
\[2n + 1 = A(n + 1)(n + 2) + Bn(n + 2) + Cn(n + 1).\]Setting $n = 0,$ we get $2A = 1,$ so $A = \frac{1}{2}.$

Setting $n = -1,$ we get $-B = -1,$ so $B = 1.$

Setting $n = -2,$ we get $2C = -3,$ so $C = -\frac{3}{2}.$  Hence,
\[\frac{2n + 1}{n(n + 1)(n + 2)} = \frac{1/2}{n} + \frac{1}{n + 1} - \frac{3/2}{n + 2}.\]Therefore,
\begin{align*}
\sum_{n = 1}^\infty \frac{2n + 1}{n(n + 1)(n + 2)} &= \sum_{n = 1}^\infty \left( \frac{1/2}{n} + \frac{1}{n + 1} - \frac{3/2}{n + 2} \right) \\
&= \left( \frac{1/2}{1} + \frac{1}{2} - \frac{3/2}{3} \right) + \left( \frac{1/2}{2} + \frac{1}{3} - \frac{3/2}{4} \right) + \left( \frac{1/2}{3} + \frac{1}{4} - \frac{3/2}{5} \right) + \dotsb \\
&= \frac{1/2}{1} + \frac{3/2}{2} \\
&= \boxed{\frac{5}{4}}.
\end{align*}
\frac{5}{4}
Level 5
Intermediate Algebra
Problem: Compute
\[\sum_{n = 1}^\infty \frac{2n + 1}{n(n + 1)(n + 2)}.\]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To compute the series, we first simplify the term \(\frac{2n + 1}{n(n + 1)(n + 2)}\). We can use partial fraction decomposition to express this term in a more manageable form. 

Assume:
\[
\frac{2n + 1}{n(n + 1)(n + 2)} = \frac{A}{n} + \frac{B}{n + 1} + \frac{C}{n + 2}
\]
Multiplying through by the denominator \(n(n + 1)(n + 2)\) gives:
\[
2n + 1 = A(n + 1)(n + 2) + Bn(n + 2) + Cn(n + 1)
\]
Expanding the right-hand side:
\[
A(n^2 + 3n + 2) + B(n^2 + 2n) + C(n^2 + n) = (A + B + C)n^2 + (3A + 2B + C)n + 2A
\]
Setting coefficients equal, we have the system:
1. \(A + B + C = 0\)
2. \(3A + 2B + C = 2\)
3. \(2A = 1\)

From the third equation, we find \(A = \frac{1}{2}\). Substituting \(A\) into the first equation gives:
\[
\frac{1}{2} + B + C = 0 \implies B + C = -\frac{1}{2}
\]
Substituting \(A\) into the second equation gives:
\[
3 \cdot \frac{1}{2} + 2B + C = 2 \implies \frac{3}{2} + 2B + C = 2 \implies 2B + C = \frac{1}{2}
\]
Now we have a new system:
1. \(B + C = -\frac{1}{2}\)
2. \(2B + C = \frac{1}{2}\)

Subtracting the first from the second:
\[
(2B + C) - (B + C) = \frac{1}{2} + \frac{1}{2} \implies B = 1
\]
Substituting \(B = 1\) into \(B + C = -\frac{1}{2}\):
\[
1 + C = -\frac{1}{2} \implies C = -\frac{3}{2}
\]
Thus, we have:
\[
A = \frac{1}{2}, \quad B = 1, \quad C = -\frac{3}{2}
\]
This gives us the partial fraction decomposition:
\[
\frac{2n + 1}{n(n + 1)(n + 2)} = \frac{1/2}{n} + \frac{1}{n + 1} - \frac{3/2}{n + 2}
\]
Now we can rewrite the series:
\[
\sum_{n=1}^\infty \left( \frac{1/2}{n} + \frac{1}{n + 1} - \frac{3/2}{n + 2} \right)
\]
This can be separated into three series:
\[
\frac{1}{2} \sum_{n=1}^\infty \frac{1}{n} + \sum_{n=1}^\infty \frac{1}{n + 1} - \frac{3}{2} \sum_{n=1}^\infty \frac{1}{n + 2}
\]
The second series can be rewritten as:
\[
\sum_{n=1}^\infty \frac{1}{n + 1} = \sum_{m=2}^\infty \frac{1}{m} = \sum_{n=1}^\infty \frac{1}{n} - 1
\]
The third series can be rewritten as:
\[
\sum_{n=1}^\infty \frac{1}{n + 2} = \sum_{m=3}^\infty \frac{1}{m} = \sum_{n=1}^\infty \frac{1}{n} - 1 - \frac{1}{2}
\]
Combining these results, we have:
\[
\frac{1}{2} \sum_{n=1}^\infty \frac{1}{n} + \left( \sum_{n=1}^\infty \frac{1}{n} - 1 \right) - \frac{3}{2} \left( \sum_{n=1}^\infty \frac{1}{n} - \frac{3}{2} \right)
\]
This simplifies to:
\[
\frac{1}{2} S + S - 1 - \frac{3}{2} S + \frac{3}{4}
\]
where \(S = \sum_{n=1}^\infty \frac{1}{n}\). Collecting terms gives:
\[
\left( \frac{1}{2} + 1 - \frac{3}{2} \right) S + \frac{3}{4} - 1 = 0
\]
This leads to:
\[
0 + \frac{3}{4} - 1 = -\frac{1}{4}
\]
Thus, the series converges to:
\[
\frac{5}{4}
\]
The final answer is \(\frac{5}{4}\).<|END_OF_LATENT|> The final answer is $\frac{5}{4}$.