📂Calculus

Limit Comparison Test

Summary¹

Given two series $\sum a_{n}$ and $\sum b_{n}$, let us assume $a_{n}, b_{n} \gt 0$. If there exists a positive number $c \gt 0$ such that

$$ \lim\limits_{n \to \infty} \dfrac{a_{n}}{b_{n}} = c $$

is satisfied, then either both series converge, or both diverge.

Explanation

This is called the limit comparison test. The comparison test is intuitive and useful, but it can only determine the convergence of a series whose terms are smaller than those of a convergent series. For example, $\sum \dfrac{1}{2^{n}}$ converges since it is a geometric series with $r = \dfrac{1}{2}$, and from the comparison test, we can deduce that $\sum \dfrac{1}{2^{n} + 1}$ also converges. However, in cases like $\sum \dfrac{1}{2^{n} - 1}$ where it converges, but $\dfrac{1}{2^{n} - 1} \gt \dfrac{1}{2^{n}}$, we cannot determine its behavior using the comparison test. The limit comparison test is useful in such situations.

Proof

Assume $m$ and $M$ are positive numbers satisfying $m \lt c \lt M$. Since $\lim\limits_{n \to \infty} \dfrac{a_{n}}{b_{n}} = c$, the following holds for sufficiently large $N$:

$$ m \lt \dfrac{a_{n}}{b_{n}} \lt M \qquad \forall n \ge N $$

$$ \implies m \cdot b_{n} \lt a_{n} \lt M \cdot b_{n} \qquad \forall n \ge N $$

If $\sum b_{n}$ converges, $\sum M \cdot b_{n}$ also converges, and by the comparison test, $\sum a_{n}$ converges as well. Conversely, if $\sum b_{n}$ diverges, $\sum m \cdot b_{n}$ also diverges, and by the comparison test, $\sum a_{n}$ diverges as well.

Similarly, if $\sum a_{n}$ converges, then $\sum b_{n}$ also converges, and if $\sum a_{n}$ diverges, then $\sum b_{n}$ also diverges.

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