Comparison Test

Theorem

Let $\displaystyle \sum_{n=1}^\infty b_n$ be a convergent series of positive real numbers.

Let $\left \langle {a_n} \right \rangle$ be a sequence in $\R$.

Let $\forall n \in \N^*: \left|{a_n}\right| \le b_n$.

Then the series $\displaystyle \sum_{n=1}^\infty a_n$ converges.

Corollary

Let $\displaystyle \sum_{n=1}^\infty b_n$ be a convergent series of positive real numbers.

Let $\left \langle {a_n} \right \rangle$ be a sequence in $\R$.

Let $H \in \R$.

Let $\exists M: \forall n > M: \left|{a_n}\right| \le H b_n$.

Then the series $\displaystyle \sum_{n=1}^\infty a_n$ converges.

Proof

Let $\epsilon > 0$.

As $\displaystyle \sum_{n=1}^\infty b_n$ converges, its tail tends to zero.

So $\displaystyle \exists N: \forall n > N: \sum_{k = n+1}^\infty b_k < \epsilon$.

Let $\left \langle s_n \right \rangle$ be the sequence of partial sums of $\displaystyle \sum_{n=1}^\infty a_n$.

Then $\forall n > m > N$:

	\(\displaystyle \)	\(\displaystyle \left\vert {s_n - s_m} \right\vert\)	\(=\)	\(\displaystyle \left\vert {\left({a_1 + a_2 + \cdots + a_n}\right) - \left({a_1 + a_2 + \cdots + a_m}\right)} \right\vert\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left\vert{a_{m+1} + a_{m+2} + \cdots + a_n}\right\vert\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\le\)	\(\displaystyle \left\vert { a_{m+1} } \right\vert + \left\vert { a_{m+2} } \right\vert + \cdots + \left\vert { a_n } \right\vert\)	\(\displaystyle \)	Triangle Inequality
	\(\displaystyle \)	\(\displaystyle \)	\(\le\)	\(\displaystyle b_{m+1} + b_{m+2} + \cdots + b_n\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\le\)	\(\displaystyle \sum_{k = n+1}^\infty b_k < \epsilon\)	\(\displaystyle \)

So $\left \langle s_n \right \rangle$ is a Cauchy sequence and the result follows from Convergent Sequence is Cauchy Sequence.

$\blacksquare$

Proof of Corollary

Let $\epsilon > 0$.

Then $\frac \epsilon H > 0$.

As $\displaystyle \sum_{n=1}^\infty b_n$ converges, its tail tends to zero.

So $\displaystyle \exists N: \forall n > N: \sum_{k = n+1}^\infty b_k < \frac \epsilon H$.

Let $\left \langle s_n \right \rangle$ be the sequence of partial sums of $\displaystyle \sum_{n=1}^\infty a_n$.

Then $\forall n > m > \max \left\{{M, N}\right\}$:

	\(\displaystyle \)	\(\displaystyle \left\vert{s_n - s_m}\right\vert\)	\(=\)	\(\displaystyle \left\vert{\left({a_1 + a_2 + \cdots + a_n}\right) - \left({a_1 + a_2 + \cdots + a_m}\right)}\right\vert\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left\vert{a_{m+1} + a_{m+2} + \cdots + a_n}\right\vert\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\le\)	\(\displaystyle \left\vert { a_{m+1} } \right\vert + \left\vert { a_{m+2} } \right\vert + \cdots + \left\vert { a_n } \right\vert\)	\(\displaystyle \)	Triangle Inequality
	\(\displaystyle \)	\(\displaystyle \)	\(\le\)	\(\displaystyle H b_{m+1} + H b_{m+2} + \cdots + H b_n\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\le\)	\(\displaystyle H \sum_{k = n+1}^\infty b_k < H \frac \epsilon H\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \epsilon\)	\(\displaystyle \)

So $\left \langle s_n \right \rangle$ is a Cauchy sequence and the result follows from Convergent Sequence is Cauchy Sequence.

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 6.15$