The Limit Comparison Test for Positive Series Examples 1

# The Limit Comparison Test for Positive Series Examples 1

Recall from The Limit Comparison Test for Positive Series page that if $\{ a_n \}$ and $\{ b_n \}$ are positive sequences where $\lim_{n \to \infty} \frac{a_n}{b_n} = L$ then:

• If $0 < L < \infty$ then either both of the series $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ both converge or both diverge.
• If $L = 0$ and $\sum_{n=1}^{\infty} b_n$ converges then $\sum_{n=1}^{\infty} a_n$ converges. If $L = 0$ and $\sum_{n=1}^{\infty} b_n$ diverges, then the test is inconclusive.
• If $L = \infty$ and $\sum_{n=1}^{\infty} b_n$ diverges then $\sum_{n=1}^{\infty} a_n$ diverges. If $L = \infty$ and $\sum_{n=1}^{\infty} b_n$ converges, then the test is inconclusive.

Let's now look at some examples of applying the limit comparison test.

## Example 1

Determine whether the series $\sum_{n=1}^{\infty} \frac{2n^2 + 3n - 1}{n^3 - 2n^2 + 4}$ converges or diverges.

Notice that for $n$ sufficiently large, the series above behaves like $\sum_{n=1}^{\infty} \frac{2n^2}{n^3} = \sum_{n=1}^{\infty} \frac{2}{n}$. We note that $\sum_{n=1}^{\infty} \frac{2}{n}$ diverges as a harmonic series. Now:

(1)
\begin{align} \quad \quad \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{2n^2 + 3n - 1}{n^3 - 2n^2 + 4}}{\frac{2}{n}} = \lim_{n \to \infty} \frac{2n^3 + 3n^2 - n}{2n^3 - 4n^2 +8} = 1 \end{align}

Therefore by the limit comparison test, we have that $\sum_{n=1}^{\infty} \frac{2n^2 + 3n - 1}{n^3 - 2n^2 + 4}$ also diverges.

## Example 2

Determine whether the series $\sum_{n=1}^{\infty} \frac{4^n + 5}{7^n - 42}$ converges or diverges.

Notice that for $n$ sufficiently large, the series above behaves like $\sum_{n=1}^{\infty} \frac{4^n}{7^n} = \sum_{n=1}^{\infty} \left ( \frac{4}{7} \right )^n$ which converges as a geometric series. Now:

(2)
\begin{align} \quad \quad \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{4^n + 5}{7^n - 42}}{\frac{4^n}{7^n}} = \lim_{n \to \infty} \frac{7^n4^n + 7^n5}{4^n7^n - 4^n42} = \lim_{n \to \infty} \frac{1 + \frac{5}{4^n}}{1 - \frac{42}{7^n}} = 1 \end{align}

Therefore $\sum_{n=1}^{\infty} \frac{4^n + 5}{7^n - 42}$ also converges.