# 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)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)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)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)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.