Questions: Test the series below for convergence using the Ratio Test. sum from n=1 to infinity of (3^n)/(n!)

Solution Steps

To test the series \(\sum_{n=1}^{\infty} \frac{3^{n}}{n!}\) for convergence using the Ratio Test, we need to find the limit of the ratio of consecutive terms. Specifically, we need to compute:

\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]

where \(a_n = \frac{3^n}{n!}\). Simplifying this ratio will help us determine the convergence of the series.

We are given the series

\[ \sum_{n=1}^{\infty} \frac{3^{n}}{n!} \]

where the \(n\)-th term is defined as

\[ a_n = \frac{3^n}{n!}. \]

To apply the Ratio Test, we need to compute the limit of the ratio of consecutive terms:

\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. \]

Calculating this gives:

\[ \frac{a_{n+1}}{a_n} = \frac{3^{n+1}/(n+1)!}{3^n/n!} = \frac{3^{n+1} \cdot n!}{3^n \cdot (n+1)!} = \frac{3^{n+1}}{3^n} \cdot \frac{n!}{(n+1)!} = \frac{3}{n+1}. \]

Now we evaluate the limit:

\[ \lim_{n \to \infty} \frac{3}{n+1} = 0. \]

Since the limit is \(0\), which is less than \(1\), the Ratio Test indicates that the series converges.

Final Answer