How do you use the limit comparison test to determine if #Sigma (n+3)/(n(n+2))# from #[1,oo)# is convergent or divergent?

Note that:

Then:

and as the harmonic series:

is divergent, by direct comparison also the series:

is divergent.

To use the limit comparison test, you select a known series that behaves similarly to the given series. In this case, we'll compare the series ( \sum_{n=1}^{\infty} \frac{n+3}{n(n+2)} ) to a known series ( \sum_{n=1}^{\infty} \frac{1}{n} ).

Then, compute the limit:

[ \lim_{n \to \infty} \frac{(n+3)/(n(n+2))}{1/n} ]

If this limit equals a finite, positive number, then both series converge or diverge together. If it equals zero, the given series converges. If it's infinity or undefined, the given series diverges.

Let's compute the limit:

[ \lim_{n \to \infty} \frac{(n+3)/(n(n+2))}{1/n} = \lim_{n \to \infty} \frac{(n+3)(n)}{n(n+2)} ]

[ = \lim_{n \to \infty} \frac{n^2 + 3n}{n^2 + 2n} = \lim_{n \to \infty} \frac{1 + 3/n}{1 + 2/n} ]

[ = \frac{1}{1} = 1 ]

Since the limit is a finite positive number, the given series ( \sum_{n=1}^{\infty} \frac{n+3}{n(n+2)} ) behaves similarly to the known series ( \sum_{n=1}^{\infty} \frac{1}{n} ). Therefore, both series either converge or diverge together. As the known series ( \sum_{n=1}^{\infty} \frac{1}{n} ) diverges (harmonic series), by the limit comparison test, the given series ( \sum_{n=1}^{\infty} \frac{n+3}{n(n+2)} ) also diverges.