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

Question

Paisley Johnson · Accepted Answer

The infinite series converges (see below).

Let #a_{n}=(2n^2)/(n!)#. Then #|a_{n+1}|/|a_{n}|=(2(n+1)^2)/((n+1)!) * (n!)/(2n^2)=((n+1)^2)/(n^2(n+1))=(n+1)/n^2# for all positive integers #n#. Since #lim_{n->infty}|a_{n+1}|/|a_{n}|=lim_{n->infty}(n+1)/n^2=0<1#, it follows that the infinite series #sum_{n=1}^{infty}(2n^2)/(n!)# converges by the Ratio Test .

Emma Aldridge · Answer

undefined 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.