# 11.6.8 determine convergent or divergence by Ratio Test

• MHB
• karush
In summary, the Ratio Test can be used to determine whether a series is convergent or divergent. If the limit of the absolute value of the ratio of consecutive terms is greater than 1 or infinity, then the series is divergent. Additionally, when applying the Ratio Test to the series $\sum_{n=1}^{\infty}\dfrac{(-2)^n}{n^2}$ and $\sum_{n=1}^{\infty} \frac{(2n)!}{9^n n!}$, it is found that both series are divergent.
karush
Gold Member
MHB
Use the Ratio Test to determine whether the series is convergent or divergent
$$\sum_{n=1}^{\infty}\dfrac{(-2)^n}{n^2}$$
If $\displaystyle\lim_{n \to \infty} \left|\dfrac{a_{n+1}}{a_n}\right|=L>1 \textit{ or } \left|\dfrac{a_{n+1}}{a_n}\right|=\infty \textit{ then the series } \sum_{n=1}^{\infty}a_n \textit{ is divergent}$

ok I spent about half hour trying to do this but not sure what a is

also this one if can...

$$\sum_{n=1}^{\infty} \frac{(2n)!}{9^n n!}$$

karush said:
Use the Ratio Test to determine whether the series is convergent or divergent
$$\sum_{n=1}^{\infty}\dfrac{(-2)^n}{n^2}$$
If $\displaystyle\lim_{n \to \infty} \left|\dfrac{a_{n+1}}{a_n}\right|=L>1 \textit{ or } \left|\dfrac{a_{n+1}}{a_n}\right|=\infty \textit{ then the series } \sum_{n=1}^{\infty}a_n \textit{ is divergent}$

ok I spent about half hour trying to do this but not sure what a is

also this one if can...

$$\sum_{n=1}^{\infty} \frac{(2n)!}{9^n n!}$$

$a_{n+1} = \dfrac{(-2)^{n+1}}{(n+1)^2}$
$a_n = \dfrac{(-2)^n}{n^2}$

$\dfrac{a_{n+1}}{a_n} = \dfrac{(-2)^{n+1}}{(n+1)^2} \cdot \dfrac{n^2}{(-2)^n} = \dfrac{-2n^2}{(n+1)^2}$

$$\displaystyle \lim_{n \to \infty} \bigg|\dfrac{-2n^2}{(n+1)^2} \bigg| = \lim_{n \to \infty} \dfrac{2}{1 + \frac{2}{n} + \frac{1}{n^2}}$$

so ... what next?

-----------------------------------------------------------------------------

$a_{n+1} = \dfrac{[2(n+1)]!}{9^{n+1} \cdot (n+1)!}$

$a_n = \dfrac{(2n)!}{9^n \cdot n!}$$$\displaystyle \lim_{n \to \infty} \bigg|\dfrac{[2(n+1)]!}{9^{n+1} \cdot (n+1)!} \cdot \dfrac{9^n \cdot n!}{(2n)!} \bigg| =$$

$$\displaystyle \lim_{n \to \infty} \bigg|\dfrac{(2n+2)!}{9^{n+1} \cdot (n+1)!} \cdot \dfrac{9^n \cdot n!}{(2n)!} \bigg|$$

$$\displaystyle \lim_{n \to \infty} \bigg|\dfrac{(2n+2)(2n+1)}{9 \cdot (n+1)} \bigg|$$

now what ... ?

$\displaystyle\lim_{n \to \infty} \dfrac{2}{1 + \frac{2}{n} + \frac{1}{n^2}}=\dfrac{2}{1+0+0}=2$
so $L>1$ divergent.and$\displaystyle\lim_{n \to \infty} \bigg|\dfrac{(2n+2)(2n+1)}{9 \cdot (n+1)} \bigg| =\bigg|\dfrac{2(n+1)(2n+1)}{9(n+1)}\bigg|=\dfrac{4n+2}{9}=\infty$
so $L=\infty$ divergent.

Last edited:
Mahalo much...

## 1. What is the Ratio Test used for?

The Ratio Test is used to determine whether an infinite series converges or diverges. It is particularly useful for series with terms involving factorials or exponential functions.

## 2. How does the Ratio Test work?

The Ratio Test compares the absolute value of consecutive terms in a series. If the limit of the ratio of these terms is less than 1, the series converges. If the limit is greater than 1, the series diverges. If the limit is exactly 1, the test is inconclusive and another method must be used.

## 3. When should I use the Ratio Test?

The Ratio Test is most effective when the terms in a series involve factorials or exponential functions. It is also useful for series with alternating signs.

## 4. Can the Ratio Test be used for all infinite series?

No, the Ratio Test can only be used for series with positive terms. It also cannot be used for series with terms that do not approach 0 as n approaches infinity.

## 5. Are there any limitations to the Ratio Test?

Yes, the Ratio Test can only determine convergence or divergence, it cannot determine the exact value of a convergent series. It also cannot be used to compare the convergence of two different series.

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