Convergence of sequences
(The following problem is borrowed from page 599 of Calculus (8th ed.) by Larson, Hostetler, and Edwards.)Problem:
Find a definition for the sequence $\{a_n\}$ whose first five terms are $$-\frac{2}{1},\frac{8}{2},-\frac{26}{6},\frac{80}{24},-\frac{242}{120},\dots$$and determine whether it converges or diverges.
Solution:
By inspection,
$$a_n = (-1)^n\frac{3^n -1}{n!}$$A theorem states that if the absolute ratio of successive terms in a sequence is less than $1$ for large $n$, then the sequence will converge to $0$, i.e.
(Note that this is similar to, but different from, the ratio test for convergence of series.)In our example,
$$\lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty}\left|\frac{(-1)^{n+1}\frac{3^{n+1}-1}{(n+1)!}}{(-1)^n\frac{3^n-1}{n!}} \right|$$We can simply ignore the $(-1)^n$ factors, because we are only interested in the ratio of the absolute values. Simplifying, we have
$$\lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n}\right| =\lim_{n\to\infty}\frac{n!}{(n+1)!}\frac{3^{n+1}-1}{3^n -1} = \lim_{n\to\infty}\frac{1}{n+1}\frac{3^{n+1}-1}{3^n -1}$$As $n\to\infty$, the first fraction in this limit goes to $0$, while the second goes to $3$. Since the limit of a product is the same as the product of the limits, the overall limit is $0$, and by the theorem stated above, the sequence converges to $0$, i.e.
$$\lim_{n\to\infty} a_n = 0$$