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 $y: \; x \lt L \lt y$. Now for large $n$, the ratio $a_n / b_n$ is very close to $L$. Another way to say that is that there must be an integer $N$ such that

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

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 $\Sigma b_n$ converges, so does $\Sigma y b_n$, because $\Sigma y b_n = y \Sigma b_n$ (properties of summations). That's just part 1 of the comparison test.

By the same logic, if $\Sigma b_n$ *diverges*, so does $\Sigma x b_n$, because $\Sigma x b_n = x \Sigma b_n$. This is just part 2 of the comparison test — the test for divergence.

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

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