📂Calculus

The Comparison Test

Theorem¹

For two series $\sum a_{n}$ and $\sum b_{n}$, suppose $a_{n}, b_{n} \gt 0$. Then the following hold.

1. If $\forall n \ a_{n} \le b_{n}$ and $\sum b_{n}$ converges, then $\sum a_{n}$ also converges.
2. If $\forall n \ a_{n} \ge b_{n}$ and $\sum b_{n}$ diverges, then $\sum a_{n}$ also diverges.

Explanation

This is called the (direct) comparison test. It is usually called simply the comparison test, but since there is a similarly named limit comparison test, it is sometimes called the direct comparison test when a clear distinction is needed.

This is an intuitive theorem stating that if a larger series converges, then a smaller series also converges, and if a smaller series diverges, then a larger series also diverges. When determining the convergence of a series with the comparison test, the following two series are frequently used.

• $p$-series $\sum \dfrac{1}{n^{p}}$

• geometric series $\sum ar^{n}$

Proof

First, let us adopt the following notation.

$$ s_{n} = \sum\limits_{i = 1}^{n} a_{i}, \qquad t_{n} = \sum\limits_{i = 1}^{n} b_{i}, \qquad t = \sum\limits_{i = 1}^{\infty} b_{i} $$

1.

Since each term of both given series is greater than $0$, $\left\{ s_{n} \right\}$ and $\left\{ t_{n} \right\}$ are increasing sequences. For all $n$, $t_{n} \le t$. By assumption $s_{n} \le t_{n}$, so

$$ s_{n} \le t \quad \forall n $$

Since $s_{n}$ is a bounded increasing sequence, by the monotone sequence theorem, $s_{n}$ converges.

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2.

If $\sum b_{n}$ diverges, then since $\left\{ t_{n} \right\}$ is an increasing sequence, $\lim\limits_{n \to \infty}t_{n} = \infty$. Since $s_{n} \ge t_{n}$, by the definition of divergence of a sequence, $\lim\limits_{n \to \infty} s_{n} = \sum a_{n} = \infty$.

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