The **limit comparison test** is an easy way to compare the limit of the terms of one series with the limit of terms of a known series to check for convergence or divergence. It may be one of the most useful tests for convergence.

The limit comparison test (**LCT**) differs from the direct comparison test. In the comparison test, we compare series elements term-by-term. In the LCT we compare the limits on the sizes of the terms as **n → ∞**.

Suppose that $\sum a_n$ and $\sum b_n$ are series with positive terms.

$$\text{If } \; \lim_{n\to\infty} \, \frac{a_n}{b_n} = L$$

and $L \gt 0,$ then either both series converge or they both diverge.

Let **x** and **y** be positive numbers and let's further say that **L** is between **x** and **y: x < L < y**. Now for large **n**, the ratio **a _{n}/b_{n}** is very close to

$$x \lt \frac{a_n}{b_n} \lt y \; \text{ when } \; n \gt N$$

Now if we multiply that inequality through by **b _{n}**, we get

$$x b_n \lt a_n \lt yb_n \; \text{ when } \; n \gt N$$

Now if **Σb _{n}** converges, so does

By the same logic, if **Σb _{n}**

** Solution**: It's reasonable to suspect that this series converges because it's almost a

$$\sum_{n = 1}^{\infty} \frac{1}{3^n}$$

We set up the limit like this. In the second step we multiply by the reciprocal of the denominator (the rules of basic algebra never change!):

$$\lim_{n\to\infty} \frac{\frac{1}{3^n - 1}}{\frac{1}{3^n}} = \lim_{n\to\infty} \frac{3^n}{3^n - 1}$$

Now to get a better look at that limit, divide every term by **3 ^{n}**:

$$= \lim_{n\to\infty} \frac{3^n/3^n}{3^n/3^n - 1/3^n}$$

which reduces to

$$= \lim_{n\to\infty} \frac{1}{1 - 1/3^n}= 1 \gt 0$$

Now this limit is easy to evaluate. As **n → ∞** the fraction equals 1, which is greater than zero, so the series converges by limit comparison with a known convergent series.

Does $\sum_{n = 1}^{\infty} \frac{n + 1}{n \sqrt{n}}$ converge?

** Solution**: The first thing we need to do in such problems is to find some approximation of the series. For large

$$\frac{n + 1}{n \sqrt{n}} \approx \frac{1}{\sqrt{n}}$$

Here's the limit expression. It reduces nicely to an easy-to-evaluate limit:

$$\lim_{n\to\infty} \frac{n + 1}{n \sqrt{n}} \frac{\sqrt{n}}{1} = \lim_{n\to\infty} \frac{n + 1}{n}$$

By dividing everything by **n**, we get to a limit that's easier to see:

$$= \lim_{n\to\infty} 1 + \frac{1}{n} = 1 \gt 0$$

The limit is greater than zero, and we compared our series to a divergent series, therefore the series we tested is divergent.

This series could also have been compared directly by asking whether its terms are, term-by-term, greater than those of the divergent series with terms **1/n ^{1/2}**.

$$\frac{n + 1}{n\sqrt{n}} \; \gt \, ? \; \frac{1}{\sqrt{n}}$$

Cross multiplication gives

$$n \sqrt{n} + \sqrt{n} \; \gt \; n \sqrt{n}$$

Subtracting terms from both sides gives

$$\sqrt{n} \gt 0$$

which is true for all **n ≥ 1**, as defined in the series.

Determine whether these series converge using the limit comparison test (LCT).

1. | $$\sum_{k = 1}^{\infty} \frac{\sqrt{2n + 1}}{n^3 - 4}$$ | |

2. | $$\sum_{n = 2}^{\infty} \frac{n^3 - 2n}{n^4 + 3}$$ | |

3. | $$\sum_{n = 1}^{\infty} \frac{n}{\sqrt{n^2 + 1}}$$ |

4. | $$\sum_{n = 3}^{\infty} \frac{3}{\sqrt{n^2 - 4}}$$ | |

5. | $$\sum_{n = 1}^{\infty} \frac{n}{(n + 1) 2^{n - 1}}$$ | |

6. | $$\sum_{n = 1}^{\infty} sin \left( \frac{1}{n} \right)$$ |

**xaktly.com** by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.© 2012-2019, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.