3.7 Limit Comparison Test

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We will use the LCT to determine if a series converges or diverges.

Limit Comparison Test

We have seen that the Direct Comparison Test can be inconclusive if the comparison goes in the wrong direction. In these cases, the Limit Comparison Test (LCT) can be used instead. As its name suggests, the LCT involves computing a limit. More precisely, it involves computing the limit of the ratio of a given series with another series whose behavior is known. The idea is that if the limit of the ratio of these two series is a positive number, $L$ , then the two series will have the same behavior, as one of them is essentially a multiple of the other. We state this in the following theorem.

Limit Comparison Test Suppose $a_n, b_n > 0$ for all $n$ . If $\lim _{n \to \infty } \frac {a_n}{b_n} = L$ where $0<L<\infty ,$ then the two infinite series $\sum _{n=1}^\infty a_n \;\; \text { and } \;\; \sum _{n=1}^\infty b_n$ have the same behavior, i.e., they either both converge or both diverge.

In practical applications of the LCT, the given series is $\displaystyle {\sum _{n=1}^\infty a_n}$ and the series we choose to compare it with is $\displaystyle {\sum _{n=1}^\infty b_n}$ .

The LCT is inconclusive if the limit DNE, but if the limit is zero or infinity, then there is a partial conclusion as mentioned in the following remark.

If $L = 0$ and $\displaystyle {\sum _{n=1}^\infty b_n}$ converges, then so does $\displaystyle {\sum _{n=1}^\infty a_n}$ and if $L = \infty$ and $\displaystyle {\sum _{n=1}^\infty b_n}$ diverges, then so does $\displaystyle {\sum _{n=1}^\infty a_n}$ .

example 1 Determine if the series $\sum _{n=1}^\infty \frac {n+2}{n^3}$ converges or diverges.(We saw this series in the section on the DCT and were not able to make a conclusion).
We will use the LCT with the convergent p-series: $\displaystyle {\sum _{n=1}^\infty \frac {1}{n^2}} \;\;(p = 2 >1)$ . We have: $\lim _{n \to \infty } \frac {\left (\frac {n+2}{n^3}\right )}{\left (\frac {1}{n^2}\right )} = \lim _{n \to \infty } \frac {(n+2)n^2}{n^3}$ $= \lim _{n \to \infty } \frac {n^3 +2n^2}{n^3} = 1,$ using the ratio of the leading coefficients. We now make our conclusion using the value $L = 1$ . Since $0 < 1 < \infty$ and the series $\displaystyle {\sum _{n=1}^\infty \frac {1}{n^2}}$ converges, the series $\displaystyle {\sum _{n=1}^\infty \frac {n+2}{n^3}}$ also converges by the LCT.

(problem 1) Determine if the series converges or diverges: $\displaystyle {\sum _{n=1}^\infty \frac {5n^2 + 1}{3n^4 - 2}}$
Which series should we compare this to?

$\displaystyle {\sum _{n=1}^\infty \frac {1}{n}}$ $\displaystyle {\sum _{n=1}^\infty \frac {1}{n^2}}$ $\displaystyle {\sum _{n=1}^\infty \frac {1}{n^3}}$

What is the value of the limit in the LCT?

$L = 0$ $L = 2$ $L = 5/3$

Describe the behavior of the series $\displaystyle {\sum _{n=1}^\infty \frac {5n^2 + 1}{3n^4 - 2}:}$

Converges by LCT Diverges by LCT No Conclusion from LCT

example 2 Determine if the series $\sum _{n=0}^\infty \frac {2^n + 4}{3^n + 1}$ converges or diverges.
We will use the LCT with the convergent geometric series: $\displaystyle {\sum _{n=1}^\infty \left (\frac 23\right )^n \;\;(r = 2/3, |r| < 1)}$ . We have: $\begin{align*} \lim_{n \to \infty} \frac{\left(\frac{2^n + 4}{3^n + 1}\right)}{\left(\frac23\right)^n} &= \lim_{n \to \infty} \frac{\left(\frac{2^n + 4}{3^n + 1}\right)}{\left(\frac{2^n}{3^n}\right)}\\ &=\lim_{n \to \infty} \frac{(2^n + 4)3^n}{(3^n + 1)2^n}\\ &= \lim_{n \to \infty} \frac{2^n + 4}{2^n} \cdot \lim_{n \to \infty}\frac{3^n}{3^n + 1}\\ &= 1 \cdot 1 \\ &= 1 \end{align*}$

using L’Hopital’s rule on each of the last two limits. We now make our conclusion using the value $L = 1$ . Since $0 < 1 < \infty$ and the series $\displaystyle {\sum _{n=0}^\infty \left (\frac 23\right )^n}$ converges, the series $\displaystyle {\sum _{n=0}^\infty \frac {2^n + 4}{3^n + 1}}$ also converges by the LCT.

(problem 2) Determine if the series converges or diverges: $\displaystyle {\sum _{n=0}^\infty \frac {5^n + 1}{2^n + 3}}$
Which series should we compare this to?

$\displaystyle {\sum _{n=0}^\infty \left (\frac 13\right )^n}$ $\displaystyle {\sum _{n=0}^\infty \left (\frac 53\right )^n}$ $\displaystyle {\sum _{n=0}^\infty \left (\frac 52\right )^n}$

What is the value of the limit in the LCT?

$L = 0$ $L = 1$ $L = 2$

Describe the behavior of the series $\displaystyle {\sum _{n=0}^\infty \frac {5^n + 1}{2^n + 3}:}$

Converges by LCT Diverges by LCT No Conclusion from LCT

example 3 Determine if the series $\sum _{n=1}^\infty \frac {n^2 + 5n + 12}{n^3 + 6n^2 + 2n}$ converges or diverges.
We will use the LCT with the divergent harmonic series: $\displaystyle {\sum _{n=1}^\infty \frac {1}{n}}$ . We have: $\lim _{n \to \infty } \frac {\left (\frac {n^2 + 5n + 12}{n^3 + 6n^2 + 2n}\right )}{\left (\frac {1}{n}\right )} = \lim _{n \to \infty } \frac {(n^2 + 5n + 12)n}{n^3 + 6n^2 + 2n}$ $= \lim _{n \to \infty } \frac {n^3 + 5n^2 + 12n}{n^3 + 6n^2 + 2n} = 1,$ using the ratio of the leading coefficients. We now make our conclusion using the value $L = 1$ . Since $0 < 1 < \infty$ and the series $\displaystyle {\sum _{n=1}^\infty \frac {1}{n}}$ diverges, the series $\displaystyle {\sum _{n=1}^\infty \frac {n^2 + 5n + 12}{n^3 + 6n^2 + 2n}}$ also diverges by the LCT.

(problem 3) Determine if the series converges or diverges: $\displaystyle {\sum _{n=1}^\infty \frac {4n^3 -3}{5n^4 + n}}$
Which series should we compare this to?

$\displaystyle {\sum _{n=1}^\infty \frac {1}{n}}$ $\displaystyle {\sum _{n=1}^\infty \frac {1}{n^2}}$ $\displaystyle {\sum _{n=1}^\infty \frac {1}{n^3}}$

What is the value of the limit in the LCT?

$L = 4/5$ $L = 5/3$ $L = 1$

Describe the behavior of the series $\displaystyle {\sum _{n=1}^\infty \frac {4n^3 -3}{5n^4 +n}:}$

Converges by LCT Diverges by LCT No Conclusion from LCT

example 4 Determine if the series $\sum _{n=1}^\infty \frac {1}{\sqrt {n^2 + 1}}$ converges or diverges.
We will use the LCT with the divergent harmonic series: $\displaystyle {\sum _{n=1}^\infty \frac {1}{n}}$ . We have: $\lim _{n \to \infty } \frac {\left (\frac {1}{\sqrt {n^2 + 1}}\right )}{\left (\frac {1}{n}\right )} = \lim _{n \to \infty } \frac {n}{\sqrt {n^2 + 1}}$ $= \lim _{n \to \infty } \frac {n}{n\sqrt {1 + \frac {1}{n^2}}} = \lim _{n \to \infty } \frac {1}{\sqrt {1 + \frac {1}{n^2}}} = 1.$ We now make our conclusion using the value $L = 1$ . Since $0 < 1 < \infty$ and the series $\displaystyle {\sum _{n=1}^\infty \frac {1}{n}}$ diverges, the series $\displaystyle {\sum _{n=1}^\infty \frac {1}{\sqrt {n^2 + 1}}}$ also diverges by the LCT.

(problem 4) Determine if the series converges or diverges: $\displaystyle {\sum _{n=1}^\infty \frac {1}{\sqrt {2n^6 - 1}}}$
Which series should we compare this to?

$\displaystyle {\sum _{n=1}^\infty \frac {1}{n^2}}$ $\displaystyle {\sum _{n=1}^\infty \frac {1}{n^3}}$ $\displaystyle {\sum _{n=1}^\infty \frac {1}{n^6}}$

What is the value of the limit in the LCT?

$L = 1$ $L = 1/2$ $L = 1/\sqrt 2$

Describe the behavior of the series $\sum _{n=1}^\infty \frac {1}{\sqrt {2n^6 - 1}}:$

Converges by LCT Diverges by LCT No Conclusion from LCT

Since the limits $\lim _{n \to \infty } \sin (n) \;\; \text { and } \;\; \lim _{n \to \infty } \cos (n)$ do not exist (due to oscillation), the LCT is generally not effective in problems involving $\sin (n)$ and $\cos (n)$ . Try the DCT instead.

Video Lesson

Here is a detailed, lecture style video on the Limit Comparison Test:

Limit Comparison Test

Video Lesson

More Problems