The Ratio Test for Positive Series Examples 2

# The Ratio Test for Positive Series Examples 2

Suppose that $\{ a_n \}$ is an ultimately positive sequence and that $\lim_{n \to \infty} \frac{a_{n+1}}{a_{n}} = \rho$ where $\rho$ is a real number or $\rho$ is $\infty$. Recall from The Ratio Test for Positive Series page that:

• If $0 ≤ \rho < 1$ then the series $\sum_{n=1}^{\infty} a_n$ is convergent.
• If $1 < \rho ≤ \infty$ then the series $\sum_{n=1}^{\infty} a_n$ is divergent.
• If $\rho = 1$ then the ratio test fails as it provides no useful information regarding the convergence or divergence of a series. ||

We will now look at some more examples applying the ratio test.

## Example 1

Using the ratio test, determine whether the series $\sum_{n=1}^{\infty} \frac{(4n)!}{4^n n!}$ converges or diverges.

This series is positive for all $n \in \mathbb{N}$ and so we can apply the ratio test. We have that:

(1)
\begin{align} \quad \lim_{n \to \infty} \frac{\frac{(4(n+1))!}{4^{n+1} (n+1)!}}{\frac{(4n)!}{4^n n!}} = \lim_{n \to \infty} \frac{(4n+4))!}{4^{n+1} (n+1)!} \frac{4^n n!}{(4n)!} = \lim_{n \to \infty} \frac{(4n + 1)(4n + 2)(4n + 3)(4n + 4)}{4 (n+1)} = \lim_{n \to \infty} (4n + 1)(4n + 2)(4n + 3) = \infty \end{align}

Therefore by the ratio test we have that $\sum_{n=1}^{\infty} \frac{(4n)!}{4^n n!}$ diverges.

## Example 2

Using the ratio test, determine whether the series $\sum_{n=1}^{\infty} \frac{n^2 + 2n + 1}{3^n + 2}$ converges or diverges.

Once again, this series is positive for all $n \in \mathbb{N}$ and so we can apply the ratio test. We have that:

(2)
\begin{align} \quad \lim_{n \to \infty} \frac{\frac{(n+1)^2 + 2(n+1) + 1}{3^{n+1} + 2}}{\frac{n^2 + 2n + 1}{3^n + 2}} = \lim_{n \to \infty} \frac{(n+1)^2 + 2(n+1) + 1}{3^{n+1} + 2} \frac{3^n + 2}{n^2 + 2n + 1} = \lim_{n \to \infty} \frac{(n^2 + 2n + 1 + 2n + 2 + 1)(3^n + 2)}{(3^{n+1} + 2)(n^2 + 2n + 1)} \\ \quad = \lim_{n \to \infty} \frac{n^2 + 4n + 4}{n^2 + 2n + 1} = \lim_{n \to \infty} \frac{\left (1 + \frac{4}{n} + \frac{4}{n^2} \right )}{\left ( 1 + \frac{1}{2n} + \frac{1}{n^2} \right )} \frac{\frac{1}{3} + \frac{2}{3^{n+1}}}{1 + \frac{2}{3^{n+1}}} = \frac{1}{3} \end{align}

Therefore by the ratio test, we have that $\sum_{n=1}^{\infty} \frac{(4n)!}{4^n n!}$ converges.

## Example 3

Show that any positive rational function comes up inconclusive when it comes to applying the ratio test. What sort of test will determine the convergence or divergence of series of rational functions?

Consider a typical rational function:

(3)
\begin{align} \quad p(n) = \frac{a_0 + a_1n + a_2n^2 + ... + a_p n^p}{b_0 + b_1n + b_2n^2 + ... + b_r n^r} \end{align}

Assume that $\{ p(n) \}$ is ultimately positive so that we can apply the ratio test. Then we have that:

(4)
\begin{align} \quad \lim_{n \to \infty} \frac{p(n+1)}{p(n)} = \lim_{n \to \infty} \frac{a_0 + a_1(n+1) + a_2(n+1)^2 + ... + a_p (n+1)^p}{b_0 + b_1(n+1) + b_2(n+1)^2 + ... + b_r (n+1)^r} \frac{b_0 + b_1n + b_2n^2 + ... + b_r n^r}{a_0 + a_1n + a_2n^2 + ... + a_p n^p} \end{align}

The degree of the polynomial in the numerator will be $p + r$ and the degree of the polynomial in the denominator will be $p + r$. Therefore the limit above is equal to $1$ and so the ratio test is inconclusive.

For positive rational functions, we can apply the limit comparison test frequently and determine the convergence or divergence from there.

## Example 4

Reprove the ratio test.

Suppose that $\{ a_n \}$ is an ultimately positive sequence and let $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \rho$.

First consider the case where $0 ≤ \rho < 1$. Choose $r \in \mathbb{R}$ such that $\rho < r < 1$. Since $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \rho$ then for some $N \in \mathbb{N}$ we have that if $n ≥ N$ then:

(5)

Therefore we have that:

(6)

Therefore we have that $a_n ≤ r^n a_N$ for all $n ≥ N$. Note that the series $\sum_{n=N+1}^{\infty} a_n$ converges as geometric series with $\mid r \mid < 1$, and so by the comparison test, we have that $\sum_{n=1}^{\infty} a_n$ converges.

Now suppose that $1 < \rho$. Choose $r \in \mathbb{R}$ such that $1 < r < \rho$. Since $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \rho$ then for some $N \in \mathbb{N}$ we have that if $n ≥ N$ then:

(7)
Note that the series $\sum_{n=N+1}^{\infty} r^3 a_N$ diverges as a geometric series with $\mid r \mid ≥ 1$. Therefore we have by the comparison test that $\sum_{n=1}^{\infty} a_n$ also diverges.