3.9 Ratio Test

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Use the Ratio Test to determine whether an infinite series converges or diverges.

Ratio Test

In this section we will determine whether an infinite series converges or diverges using the Ratio Test. The idea behind this test is to determine if the given series is behaving similarly to a geometric series. For a geometric series, $\sum _{n=0}^\infty a_n = \sum _{n=0}^\infty ar^n,$ the ratio of successive terms is constant, i.e., $\frac {a_{n+1}}{a_n} = \frac {ar^{n+1}}{ar^n} = r.$ Furthermore, a geometric series converges if this ratio is between $-1$ and $1$ . For a general series, the ratio of successive terms is not constant, but if the limit of the ratio of successive terms approaches a value, $L$ , between $-1$ and $1$ , then the series behaves like a geometric series with ratio equal to $L$ . We state this precisely in the following theorem.

Ratio Test For the series $\sum _{n=0}^\infty a_n,$ let $L = \lim _{n \to \infty } \left |\frac {a_{n+1}}{a_n}\right |.$ The series converges (absolutely) if $L < 1$ and diverges if $L>1$ . If $L = 1,$ then no conclusion can be made.

For any $p$ -series, $\sum _{n=1}^\infty \frac {1}{n^p},$ $L = \lim _{n \to \infty } \left |\frac {\frac {1}{(n+1)^p}}{\frac {1}{n^p}}\right | = \lim _{n \to \infty } \frac {n^p}{(n+1)^p} = \lim _{n \to \infty } \left (\frac {n}{n+1}\right )^p = 1$ and so the ratio test gives no conclusion about $p$ -series.

example 1 Determine whether the series converges or diverges: $\sum _{n=1}^\infty \frac {n}{2^n}$

By the Ratio Test, $L = \lim _{n \to \infty } \left |\frac {\frac {n+1}{2^{n+1}}}{\frac {n}{2^n}}\right | = \lim _{n \to \infty } \frac {(n+1)2^n}{n2^{n+1}} = \lim _{n \to \infty } \frac {n+1}{2n} = \frac 12.$ Since $L = 1/2 < 1$ , the series converges.

(problem 1a) Determine whether the series converges or diverges: $\sum _{n=1}^\infty \frac {n+2}{3^n}$ $L = \lim _{n \to \infty } \left |\frac {a_{n+1}}{a_n}\right | = \answer {1/3}$ By the ratio test, this series converges diverges

(problem 1b) Determine whether the series converges or diverges: $\sum _{n=1}^\infty \frac {2^n}{n^3}$ $L = \lim _{n \to \infty } \left |\frac {a_{n+1}}{a_n}\right | = \answer {2}$ By the ratio test, this series converges diverges

(problem 1c) Determine whether the series converges or diverges: $\sum _{n=1}^\infty \frac {n^2}{n^4 +1}$ $L = \lim _{n \to \infty } \left |\frac {a_{n+1}}{a_n}\right | = \answer {1}$ By the ratio test, this series converges divergesno conclusion

example 2 Determine whether the series converges or diverges: $\sum _{n=1}^\infty (-1)^n\frac {n^2}{3^n}$

By the Ratio Test, $L = \lim _{n \to \infty } \left |\frac {(-1)^{n+1}\frac {(n+1)^2}{3^{n+1}}}{(-1)^n\frac {n^2}{3^n}}\right | = \lim _{n \to \infty } \frac {(n+1)^2 3^n}{n^2 3^{n+1}} = \lim _{n \to \infty } \frac {(n+1)^2}{3n^2} = \frac 13.$ Since $L = 1/3 < 1$ , the series converges.

(problem 2a) Determine whether the series converges or diverges: $\sum _{n=1}^\infty (-1)^n\frac {n}{2^n}$ $L = \lim _{n \to \infty } \left |\frac {a_{n+1}}{a_n}\right | = \answer {1/2}$ By the ratio test, this series converges diverges

(problem 2b) Determine whether the series converges or diverges: $\sum _{n=1}^\infty (-1)^n\frac {3^n}{2n+1}$ $L = \lim _{n \to \infty } \left |\frac {a_{n+1}}{a_n}\right | = \answer {3}$ By the ratio test, this series converges diverges

example 3 Determine whether the series converges or diverges: $\sum _{n=0}^\infty \frac {1}{n!}$

By the Ratio Test, $L = \lim _{n \to \infty } \left |\frac {\frac {1}{(n+1)!}}{\frac {1}{n!}}\right | = \lim _{n \to \infty } \frac {n!}{(n+1)!} = \lim _{n \to \infty } \frac {1}{n+1} = 0.$ Since $L = 0 < 1$ , the series converges.

(problem 3a) Determine whether the series converges or diverges: $\sum _{n=0}^\infty \frac {2^n}{n!}$ $L = \lim _{n \to \infty } \left |\frac {a_{n+1}}{a_n}\right | = \answer {0}$ By the ratio test, this series converges diverges

(problem 3b) Determine whether the series converges or diverges: $\sum _{n=0}^\infty (-1)^n\frac {n!}{10^n}$ $L = \lim _{n \to \infty } \left |\frac {a_{n+1}}{a_n}\right | = \answer {\infty }$ By the ratio test, this series converges diverges

example 4 Determine whether the series converges or diverges: $\sum _{n=1}^\infty \frac {n^5 10^{n+1}}{(2n)!}$

By the Ratio Test, $L = \lim _{n \to \infty } \left |\frac {\frac {(n+1)^5 10^{n+2}}{(2n+2)!}}{\frac {n^5 10^{n+1}}{(2n)!}}\right | = \lim _{n \to \infty } \frac {(n+1)^5 10^{n+2}(2n)!}{n^5 10^{n+1}(2n+2)!}$ $= \lim _{n \to \infty } \frac {(n+1)^5}{n^5} \cdot \frac {10^{n+2}}{10^{n+1}} \cdot \frac {(2n)!}{(2n+2)!}$ $= \lim _{n \to \infty } \left (\frac {n+1}{n}\right )^5 \cdot 10 \cdot \frac {1}{(2n+2)(2n+1)} = 1 \cdot 10 \cdot 0 = 0.$ Since $L = 0 < 1$ , the series converges.

(problem 4) Determine whether the series converges or diverges: $\sum _{n=0}^\infty \frac {(n+1)4^n}{n!}$ $L = \lim _{n \to \infty } \left |\frac {a_{n+1}}{a_n}\right | = \answer {0}$ By the ratio test, this series converges diverges

example 5 Determine whether the series converges or diverges: $\sum _{n=1}^\infty \frac {n!}{n^n}$

By the Ratio Test, $L = \lim _{n \to \infty } \left |\frac {\frac {(n+1)!}{(n+1)^{n+1}}}{\frac {n!}{n^n}}\right | = \lim _{n \to \infty } \frac {(n+1)!}{(n+1)^{n+1}}\cdot \frac {n^n}{n!} = \lim _{n \to \infty } \frac {(n+1)!}{n!} \cdot \frac {n^n}{(n+1)^n}\cdot \frac {1}{n+1}$ $= \lim _{n \to \infty } (n+1) \cdot \left (\frac {n}{n+1}\right )^n \cdot \frac {1}{n+1} = \lim _{n \to \infty } \left (\frac {n}{n+1}\right )^n = \frac {1}{e}.$

Since $L = 1/e < 1$ , the series converges.

We used the famous limit: $\lim _{n \to \infty } \left (\frac {n+1}{n}\right )^n = \lim _{n \to \infty } \left (1 + \frac {1}{n}\right )^n = e.$

(problem 5) Determine whether the series converges or diverges: $\sum _{n=1}^\infty \frac {(n!)^2}{(2n)!}$ $L = \lim _{n \to \infty } \left |\frac {a_{n+1}}{a_n}\right | = \answer {1/4}$ By the ratio test, this series converges diverges

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