To determine the convergence or divergence of the series S = ∑n=1∞ 1/(n² + 5n + 6), we first need to analyze the behavior of the terms of the series as n approaches infinity.
Step 1: Simplifying the Expression
The denominator can be factored as follows:
n² + 5n + 6 = (n + 2)(n + 3)
So, we can rewrite the series:
S = ∑n=1∞ 1/((n + 2)(n + 3))
Step 2: Analyzing Convergence
To analyze the convergence of this series, we can use the Comparison Test. We need to compare our series with a known convergent series. For large values of n, the term 1/((n + 2)(n + 3)) behaves similarly to 1/n² since:
1/((n + 2)(n + 3)) ≈ 1/n² as n → ∞
The series S = ∑n=1∞ 1/n² is a known convergent p-series with p = 2 (where p > 1, hence converges). Therefore, since:
0 < 1/((n + 2)(n + 3)) < C/n² for some constant C and sufficiently large n, we can conclude that the original series converges.
Step 3: Finding the Sum
To find the exact sum of this convergent series, we can apply the method of partial fractions:
1/((n + 2)(n + 3)) = A/(n + 2) + B/(n + 3)
Solving for A and B, we multiply through by the denominator:
1 = A(n + 3) + B(n + 2)
Setting up the equations and solving gives us:
Upon finding A and B, we’ll have:
S = ∑n=1∞ [A/(n + 2) + B/(n + 3)]
Both individual fractions can be summed up using telescoping series or integral tests, but the exact sum may not resolve into a simple closed form easily. However, it can be calculated numerically or through software for a precise result.
Conclusion
The series S = ∑n=1∞ 1/(n² + 5n + 6) is convergent, and while determining the exact sum may involve additional calculations, we affirm its convergence based on the Comparison Test.