#### xaktly | Infinite series

Limit comparison test

### The limit comparison test for series convergence

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 → ∞.

#### The limit comparison test

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.

#### Proof of the LCT

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 an/bn 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 bn, we get

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

Now if Σbn converges, so does Σybn, because Σybn = yΣbn (properties of summations). That's just part 1 of the comparison test.

By the same logic, if Σbn diverges, so does Σxbn, because Σxbn = xΣbn. This is just part 2 of the comparison test – the test for divergence.

### Example 1:   LCT, convergent series

Does   $\sum_{n = 1}^{\infty} \frac{1}{3^n - 1}$ converge?

Solution: It's reasonable to suspect that this series converges because it's almost a convergent geometric series. Let's do the limit comparison test with the convergent geometric series:

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

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 3n:

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

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.

### Example 2:   LCT, divergent 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 n (in which case the 1 in the numerator doesn't matter), this series is approximately equal to the divergent p-series 1/n1/2, so we can use that for the limit comparison test, in which we'll guess that the series is divergent.

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

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/n1/2.

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

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.

### Practice problems

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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