Determine whether the series \(\sum_{n=1}^{\infty }\frac{1}{(n^{2}+5n+6)}\) is convergent or divergent. If it is convergent, find its sum.

Solution:

Given, \(\sum_{n=1}^{\infty }\frac{1}{(n^{2}+5n+6)}\)

We have to find whether the series is convergent or divergent.

We have to find the sum if the series is convergent.

\(\lim_{n \to \infty }a_{n}=0\) is not sufficient to prove the series \(\sum_{n=1}^{\infty }a_{n}\).

So, take \(a_{n}=\frac{1}{n}\)

Now, n² + 5n + 6 can be written as (n + 2) (n + 3)

\(\frac{1}{(n^{2}+5n+6)}=\frac{1}{(n+2)(n+3)}\)

= \(\frac{1}{(n+2)}-\frac{1}{(n+3)}\)

So, \(\sum_{n=1}^{\infty }\frac{1}{(n^{2}+5n+6)}=\sum_{n=1}^{N}(\frac{1}{n+2}-\frac{1}{n+3})\)

S = \(\lim_{n \to \infty }S_{n}\)

Where, Sₙ are the partial sums.

a₁ = \((\frac{1}{1+2}-\frac{1}{1+3})\\=(\frac{1}{3}-\frac{1}{4})\)

\(a_{2}=(\frac{1}{2+2}-\frac{1}{2+3})\\=(\frac{1}{4}-\frac{1}{5})\)

\(a_{3}=(\frac{1}{3+2}-\frac{1}{3+3})\\=(\frac{1}{5}-\frac{1}{6})\)

S₁ = a₁ = \(\frac{1}{3}-\frac{1}{4}\)

\(S_{2}=a_{1}+a_{2}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}=\frac{1}{3}-\frac{1}{5}\)

\(S_{3}=a_{1}+a_{2}+a_{3}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}=\frac{1}{3}-\frac{1}{6}\)

\(S_{n}=a_{1}+a_{2}+a_{3}+....+a_{n}=\frac{1}{3}-\frac{1}{n+3}\)

So, \(S=\lim_{n \to \infty }(\frac{1}{3}-\frac{1}{n+3})=\frac{1}{3}\)

Therefore, the series is convergent and its sum is 1/3.

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Determine whether the series \(\sum_{n=1}^{\infty }\frac{1}{(n^{2}+5n+6)}\) is convergent or divergent. If it is convergent, find its sum.

Summary:

The series \(\sum_{n=1}^{\infty }\frac{1}{(n^{2}+5n+6)}\) is convergent and its sum is 1/3.