Limit Comparison Test for Positive Series of Real Numbers Examples 1

# The Limit Comparison Test for Positive Series of Real Numbers Examples 1

Recall from The Limit Comparison Test for Positive Series of Real Numbers page the following test for convergence/divergence of a geometric series:

The Limit Comparison Test for Positive Series of Real Numbers

Let $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ be positive sequences of real numbers and let $\displaystyle{L = \lim_{n \to \infty} \frac{a_n}{b_n}}$.

a) If we have that $0 < L < \infty$, then we conclude that:

• The series $\displaystyle{\sum_{n=1}^{\infty} a_n}$ and $\displaystyle{\sum_{n=1}^{\infty} b_n}$ either both converge or both diverge.

b) If we have that $L = 0$ and $\displaystyle{\sum_{n=1}^{\infty} b_n}$ converges, then we conclude that:

• $\displaystyle{\sum_{n=1}^{\infty} a_n}$ converges.

c) If we have that $L = \infty$ and $\displaystyle{\sum_{n=1}^{\infty} b_n}$ diverges, then we conclude that:

• $\displaystyle{\sum_{n=1}^{\infty} a_n}$ diverges.

We will now look at some examples of applying the limit comparison test.

## Example 1

Determine whether $\displaystyle{\sum_{n=1}^{\infty} \frac{3^n + 2}{5^n - n}}$ converges or diverges.

Notice that for very large $n$ that:

(1)
\begin{align} \quad \frac{3^n + 2}{5^n - n} \approx \frac{3^n}{5^n} = \left ( \frac{3}{5} \right )^n \end{align}

We know that the series $\displaystyle{\sum_{n=1}^{\infty} \left ( \frac{3}{5} \right )^n}$ converges as a geometric series, so this series will be a good series to limit compare with. So we will use the limit comparison test between $\displaystyle{\sum_{n=1}^{\infty} \frac{3^n + 2}{5^n - n}}$ and $\displaystyle{\sum_{n=1}^{\infty} \left ( \frac{3}{5} \right )^n}$. Let:

(2)
\begin{align} \quad L = \lim_{n \to \infty} \frac{3^n + 2}{5^n - n} \cdot \frac{5^n}{3^n} = \lim_{n \to \infty} \frac{3^n5^n + 2 \cdot 5^n}{3^n5^n - n3^n} = \lim_{n \to \infty} \frac{1 + \frac{2}{3^n}}{1 - \frac{n}{5^n}} = 1 \end{align}

So by the limit comparison test, the series, since $0 < L = 1 < \infty$ and $\displaystyle{\sum_{n=1}^{\infty} \frac{3^n}{5^n}}$ converges we have that $\displaystyle{\sum_{n=1}^{\infty} \frac{3^n + 2}{5^n - n}}$ converges.

## Example 2

Determine whether $\displaystyle{\sum_{n=1}^{\infty} \frac{3n^4 - 4n^2 + 2n + 9}{17n^5 - 4n^4 - 3n^3 + 2n^2}}$ converges or diverges.

This series looked incredibly daunting, however, we will see that determining convergence/divergence is actually relatively simple. Notice that for sufficiently large $n$ we have that:

(3)
\begin{align} \quad \frac{3n^4 - 4n^2 + 2n + 9}{17n^5 - 4n^4 - 3n^3 + 2n^2} \approx \frac{3n^4}{17n^5} = \frac{3}{17n} \end{align}

The series $\displaystyle{\sum_{n=1}^{\infty} \frac{3}{17n}}$ diverges as a harmonic series. Using this series to limit compare with and we see that:

(4)
\begin{align} \quad L = \lim_{n \to \infty} \frac{3n^4 - 4n^2 + 2n + 9}{17n^5 - 4n^4 - 3n^3 + 2n^2} \cdot \frac{17n}{3} = \lim_{n \to \infty} \frac{51n^5 - 68n^3 + 34n + 153}{51n^5 - 12n^4 - 9n^3 + 6n^2} = \lim_{n \to \infty} \frac{51 - \frac{68}{n^2} + \frac{34}{n^4} + \frac{153}{n^5}}{51 - \frac{12}{n} - \frac{9}{n^2} + \frac{6}{n^3}} = 1 \end{align}

So by the limit comparison test, since $0 < L = 1 < \infty$ and since $\displaystyle{\sum_{n=1}^{\infty} \frac{3}{17n}}$ diverges, we have that $\displaystyle{\sum_{n=1}^{\infty} \frac{3n^4 - 4n^2 + 2n + 9}{17n^5 - 4n^4 - 3n^3 + 2n^2}}$ diverges.