The Limit Comparison Test for Positive Series Examples 2

# The Limit Comparison Test for Positive Series Examples 2

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 more examples of applying the limit comparison test.

## Example 1

Use the limit comparison test to determine whether the series $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n} + \sqrt{n+1}}$ converges or diverges.

Note that for large $n$, we have that:

(1)
\begin{align} \frac{1}{\sqrt{n} + \sqrt{n+1}} \approx \frac{1}{\sqrt{n} + \sqrt{n}} = \frac{1}{2\sqrt{n}} \end{align}

Using the limit comparison test we have that:

(2)
\begin{align} \quad \lim_{n \to \infty} \frac{\frac{1}{\sqrt{n} + \sqrt{n+1}}}{\frac{1}{2\sqrt{n}}} = \lim_{n \to \infty} \frac{2 \sqrt{n}}{\sqrt{n} + \sqrt{n+1}} = \lim_{n \to \infty} \frac{2}{1 + \frac{\sqrt{n + 1}}{\sqrt{n}}} = \lim_{n \to \infty} \frac{2}{1 + \sqrt{1 + \frac{1}{n}}} = 1 \end{align}

We have that $\sum_{n=1}^{\infty} \frac{1}{2\sqrt{n}}$ diverges as a p-series with $p = \frac{1}{2} ≤ 1$, and so by the limit comparison test, we have that $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n} + \sqrt{n+1}}$ also diverges.

## Example 2

Use the limit comparison test to determine whether the series $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2 + 1} - \sqrt{n^2-1}}$ converges or diverges.

We will first apply the regular comparison test. Note that $\sqrt{n^2 + 1} - \sqrt{n^2 - 1} ≤ \sqrt{n^2 + 1}$ for all $n \in \mathbb{N}$. Therefore we have that:

(3)
\begin{align} \quad \frac{1}{\sqrt{n^2 + 1} - \sqrt{n^2 - 1}} ≥ \frac{1}{\sqrt{n^2 + 1}} \end{align}

Now consider the series $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2 + 1}}$. We hope to show that this series diverges. Note that for $n$ very large, we have that:

(4)
\begin{align} \quad \frac{1}{\sqrt{n^2 + 1}} \approx \frac{1}{\sqrt{n^2}} = \frac{1}{n} \end{align}

Therefore we have that:

(5)
\begin{align} \quad \lim_{n \to \infty} \frac{\frac{1}{\sqrt{n^2 + 1}}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n}{\sqrt{n^2 + 1}} = \lim_{n \to \infty} \frac{n}{n\sqrt{1 + \frac{1}{n^2}}} = \lim_{n \to \infty} \frac{1}{\sqrt{1 + \frac{1}{n^2}}} = 1 \end{align}

Therefore by the limit comparison test, we have that since $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges as the harmonic series, then $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2 + 1}}$ diverges, and since $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2 + 1}}$ diverges, we have that by the regular comparison test that $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2 + 1} - \sqrt{n^2-1}}$ diverges.