# How do you use the ratio test to test the convergence of the series #∑ (2n^2)/(n!) # from n=1 to infinity?

The infinite series converges (see below).

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To use the ratio test for the series ( \sum \frac{2n^2}{n!} ), we compute the limit as ( n ) approaches infinity of the absolute value of the ratio of the (n+1)-th term to the nth term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{2(n+1)^2}{(n+1)!}}{\frac{2n^2}{n!}} \right| ]

After simplifying:

[ \lim_{n \to \infty} \left| \frac{(n+1)^2}{n^2} \cdot \frac{n!}{(n+1)!} \right| ]

[ = \lim_{n \to \infty} \left| \frac{(n+1)^2}{n^2} \cdot \frac{n!}{(n+1)n!} \right| ]

[ = \lim_{n \to \infty} \left| \frac{(n+1)^2}{n^2} \cdot \frac{1}{n+1} \right| ]

[ = \lim_{n \to \infty} \left| \frac{n^2 + 2n + 1}{n^2} \cdot \frac{1}{n+1} \right| ]

[ = \lim_{n \to \infty} \left| 1 + \frac{2}{n} + \frac{1}{n^2} \right| \cdot \frac{1}{n+1} ]

[ = 1 \cdot 0 = 0 ]

Since the limit is less than 1, by the ratio test, the series ( \sum \frac{2n^2}{n!} ) converges absolutely.

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