To determine whether the series \( S_n = \sum_{n=1}^{\infty} \frac{1}{n^2 + 5n + 6} \) is convergent or divergent, we first analyze the expression in the denominator.
The denominator \( n^2 + 5n + 6 \) can be factored as follows:
\( n^2 + 5n + 6 = (n + 2)(n + 3) \)
Thus, we can rewrite the series as:
\( S_n = \sum_{n=1}^{\infty} \frac{1}{(n + 2)(n + 3)} \)
To check for convergence, we can apply the comparison test. Notice that for large values of \( n \), the term \( \frac{1}{(n + 2)(n + 3)} \) behaves similarly to \( \frac{1}{n^2} \).
The series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) is a well-known convergent p-series with \( p = 2 \). Therefore, we can conclude that our original series converges because we have:
\( \frac{1}{(n + 2)(n + 3)} < \frac{1}{n^2} \text{ for sufficiently large } n. \)
Since \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) converges, by the comparison test, the series \( S_n \) also converges.
Next, we find the sum of the series. We use the method of partial fractions to decompose the fraction. We can express:
\( \frac{1}{(n + 2)(n + 3)} = \frac{A}{n + 2} + \frac{B}{n + 3} \)
Multiplying through by the denominator gives us:
\( 1 = A(n + 3) + B(n + 2) \)
Setting up equations based on values of n allows us to solve for \( A \) and \( B \):
By substituting suitable values (say, \( n = -3 \) and \( n = -2 \)), or directly comparing coefficients, we find:
\( A = 1 \text{ and } B = -1 \)
Thus, we have:
\( \frac{1}{(n + 2)(n + 3)} = \frac{1}{n + 2} – \frac{1}{n + 3} \)
Now our series becomes a telescoping series:
\( S_n = \sum_{n=1}^{\infty} \left( \frac{1}{n + 2} – \frac{1}{n + 3} \right) \)
The terms will cancel out, and we can find the limit as \( n \) approaches infinity:
Thus, the limit of the sum can be evaluated, and we find that:
\( S_n = 1 – \frac{1}{4} = \frac{3}{4}. \)
In conclusion, the series converges, and the sum is:
\( \frac{3}{4}. \)