Question

Determine the convergence or divergence of the sequence with the given \(n\) th term. If the sequence converges, find its limit. \(a_{n}=\frac{(n-2) !}{n !}\)

Short answer

The series converges, and its limit is 0.

Step-by-step solution

Identify the Sequence

The sequence given in this exercise is \(a_{n}=\frac{(n-2) !}{n !}\). Remember that a sequence is an ordered list of numbers that usually forms a distinct pattern.

Simplify the Sequence

Let's simplify the sequence using the properties of factorial. \(n ! = n(n-1)(n-2)!\). Using this property we can simplify \(a_{n}\) as \(a_{n}=\frac{1}{n(n-1)}\).

Apply the Ratio Test

Apply the ratio test to determine if the series converges or diverges. The ratio test says that given a series \(a_{n}\), if the limit as \(n\) approaches infinity of the absolute value of the ratio of \(a_{n+1}\) to \(a_{n}\) is less than 1, the series converges.

Compute the Limit

Insert \(n+1\) into the simplified sequence to obtain \(a_{n+1}\). Compute the limit as \(n\) approaches infinity for the ratio of \(a_{n+1}\) over \(a_{n}\) which yields: \(\lim_{n \to \infty} \frac{a_{n+1}}{a_{n}} = \lim_{n \to \infty} \frac{\frac{1}{(n+1)n}}{\frac{1}{n(n-1)}} = \lim_{n \to \infty} \frac{n(n-1)}{(n+1)n} = \lim_{n \to \infty} \frac{n-1}{n+1}\)

Evaluate the Limit

Evaluating this limit as \(n\) approaches infinity, we have \(\lim_{n \to \infty} \frac{n-1}{n+1} = 1\). Since the limit is 1, the series neither diverges nor converges by the ratio test.

Determine Convergence or Divergence

Since the limit is neither less than nor greater than 1, the ratio test is inconclusive, and it does not provide information about the convergence or divergence of the series. On the other hand, as \(n\) approaches infinity, \(a_{n}\) approaches 0, so the sequence converges to 0 by definition of convergence.