Ratio Test

Consider the series

\[ \sum\limits_{n = 1}^{\infty} \frac{9^n}{(-2)^{n+1}n} .\]

If you leave off the absolute value signs when you take the limit, you get:

\[ \begin{aligned} L &= \lim\limits_{n \to \infty} \frac{a_{n+1}}{a_n} \\ &= \lim\limits_{n \to \infty} \frac{\frac{9^{n+1}}{(-2)^{n+2}(n+1)} }{\frac{9^n}{(-2)^{n+1}n} } \\ &= \lim\limits_{n \to \infty} \left(\frac{9^{n+1}}{(-2)^{n+2}(n+1)} \right) \left( \frac {(-2)^{n+1}n}{9^n}\right) \\ &= \lim\limits_{n \to \infty} \frac{9 n}{-2(n+1) } \\ &= -\frac{9}{2}. \end{aligned} \]

Since

\[L = -\frac{9}{2} < 1 \]

this would seem to imply that the series converges.

However if the Ratio Test is applied correctly with the absolute value signs, then

\[ L = \left| -\frac{9}{2} \right|= \frac{9}{2} > 1, \]

so in fact the Ratio Test tells you that the series diverges.

So be careful because leaving off the absolute value signs can give the wrong answer!

Decide if the series

\[ \sum\limits_{n=1}^{\infty} \frac{(-3)^n}{(2n)!} \]

converges or diverges.

Answer:

Taking the limit to find \( L \) gives

\[ \begin{aligned} L &= \lim\limits_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \\ &= \lim\limits_{n \to \infty} \left| \frac{\frac{(-3)^{n+1}}{(2(n+1))!} }{\frac{(-3)^n}{(2n)!} } \right| \\ &= \lim\limits_{n \to \infty} \left| \frac{(-3)^{n+1}}{(2(n+1))!}\cdot \frac{(2n)!}{(-3)^n} \right| \\ &= \lim\limits_{n \to \infty} \left| \frac{-3\cdot (2n)!}{(2n+2)(2n+1)(2n)! } \right| \\ &= \lim\limits_{n \to \infty} \left| \frac{-3}{(2n+2)(2n+1)} \right| \\ &= 0. \end{aligned} \]

So by the Ratio Test, this series converges.

Decide if the series

\[ \sum\limits_{n=1}^{\infty} \frac{n!}{4^n} \]

converges or diverges.

Answer:

Finding \( L \) to try to apply the Ratio Test gives you

\[ \begin{aligned} L &= \lim\limits_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \\ &= \lim\limits_{n \to \infty} \left| \frac{\frac{(n+1)!}{4^{n+1}} }{\frac{n!}{4^n} } \right| \\ &= \lim\limits_{n \to \infty} \left| \frac{(n+1)!}{4^{n+1}} \cdot \frac{4^n}{n!} \right| \\ &= \lim\limits_{n \to \infty} \left| \frac{n!(n+1)}{4\cdot n!} \right| \\ &= \lim\limits_{n \to \infty} \left| \frac{n+1}{4} \right| \\ &= \infty. \end{aligned} \]

In other words, \( L \) is undoubtedly bigger than 1, so the series diverges.

Try to apply the Ratio Test to the series

\[ \sum\limits_{n=1}^{\infty} \frac{1}{n}. \]

Answer:

You already know that this is the harmonic series, which is a P-series with \( p = 1 \), and so it diverges. But if you try to apply the Ratio Test,

\[ \begin{aligned} L &= \lim\limits_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \\ &= \lim\limits_{n \to \infty} \left| \frac{\frac{1}{n+1} }{\frac{1}{n} } \right| \\ &= \lim\limits_{n \to \infty} \frac{n}{n+1 } \\ &= 1. \end{aligned} \]

Since \( L = 1 \) the Ratio Test can't be applied to show that this series diverges.

Try to apply the Ratio Test to the series

\[ \sum\limits_{n=1}^{\infty} \frac{1}{n^2}. \]

Answer:

This is a P-series with \( p =2 \), so you know it is absolutely convergent. But can the Ratio Test tell you that? Taking the limit,

\[ \begin{aligned} L &= \lim\limits_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \\ &= \lim\limits_{n \to \infty} \left| \frac{\frac{1}{(n+1)^2} }{\frac{1}{n^2} } \right| \\ &= \lim\limits_{n \to \infty} \frac{n^2}{(n+1)^2 } \\ &= 1. \end{aligned} \]

Since \( L = 1 \) the Ratio Test can't be applied to show that this series is absolutely convergent.

Try to apply the Ratio Test to the series

\[ \sum\limits_{n=1}^{\infty} \frac{(-1)^n}{n}. \]

Answer:

This is the alternating harmonic series, so you know it is conditionally convergent. What can the Ratio Test tell you? Taking the limit,

\[ \begin{aligned} L &= \lim\limits_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \\ &= \lim\limits_{n \to \infty} \left| \frac{\frac{(-1)^{n+1}}{n+1} }{\frac{(-1)^n}{n} } \right| \\ &= \lim\limits_{n \to \infty} \frac{n}{n+1 } \\ &= 1. \end{aligned} \]

Since \( L = 1 \) the Ratio Test can't be applied to show that this series is conditionally convergent.

Determine if the series

\[ \sum\limits_{n=1}^{\infty} \frac{(-1)^n}{n^2+1} \]

is convergent or divergent.

Answer:

First let's find \( L \) and see if the Ratio Test can give you a result. Taking the limit,

\[ \begin{aligned} L &= \lim\limits_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \\ &= \lim\limits_{n \to \infty} \left| \frac{\frac{(-1)^{n+1}}{(n+1)^2 + 1} }{\frac{(-1)^n}{n^2+1} } \right| \\ &= \lim\limits_{n \to \infty} \frac{n^2+1}{(n+1)^2 + 1 } \\ &= 1, \end{aligned} \]

so in fact the Ratio Test doesn't tell you anything.

Since this is an alternating series, you can check to be sure the conditions of the Alternating Series Test are met. For this series

\[ b_n = \frac{1}{n^2+1}. \]

Note that \( b_n > 0\) so the first condition of the test is met. Also,

\[ \lim\limits_{n \to \infty} b_n = \lim\limits_{n \to \infty} \frac{1}{n^2+1} = 0, \]

and the second condition of the test is met. Checking the third condition, since

\[ (n+1)^2 + 1 > n^2 + 1, \]

you know that

\[ b_n = \frac{1}{n^2+1} < \frac{1}{(n+1)^2+1} = b_{n+1} \]

and the third condition is met as well. Therefore by the Alternating Series Test, the series is convergent.

Determine if the series

\[ \sum\limits_{n=1}^{\infty} \frac{3n+7}{2n-1} \]

is convergent or divergent.

Answer:

If you try and apply the Ratio Test, taking the limit you get

\[ \begin{aligned} L &= \lim\limits_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \\ &= \lim\limits_{n \to \infty} \left| \frac{\frac{3(n+1) +7}{2(n+1)-1} }{\frac{3n+7}{2n-1} } \right| \\ &= \lim\limits_{n \to \infty} \frac{(2n-1)(3n+10)}{(2n+1)(3n+7) } \\ &= 1, \end{aligned} \]

which means the Ratio Test can't be applied.

If instead you look at

\[ \lim\limits_{n \to \infty} \frac{3n+7}{2n-1} = \frac{3}{2}, \]

the n^th Term Test for Divergence tells you that the series diverges.