Comparison Test

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Theorem

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

Let $\sequence {a_n}$ be a sequence $\R$ or sequence in $\C$.


Let $\forall n \in \N_{>0}: \cmod {a_n} \le b_n$.


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


Corollary 1

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

Let $\sequence {a_n}$ be a sequence in $\R$ or a sequence in $\C$.

Let $H \in \R$.

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


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


Corollary 2

Let $\ds \sum_{n \mathop = 1}^\infty a_n$ and $\ds \sum_{n \mathop = 1}^\infty b_n$ be two series of positive real numbers.

Let $\lim_{n \mathop \to \infty} \dfrac {a_n} {b_n} = k$ for some $k \in \R$.


Then either:

both $\ds \sum_{n \mathop = 1}^\infty a_n$ and $\ds \sum_{n \mathop = 1}^\infty b_n$ are convergent

or:

both $\ds \sum_{n \mathop = 1}^\infty a_n$ and $\ds \sum_{n \mathop = 1}^\infty b_n$ are divergent.


Proof

Let $\epsilon > 0$.

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

So:

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


Let $\sequence {s_n}$ be the sequence of partial sums of $\ds \sum_{n \mathop = 1}^\infty a_n$.

Then $\forall n > m > N$:

\(\ds \cmod {s_n - s_m}\) \(=\) \(\ds \cmod {\paren {a_1 + a_2 + \cdots + a_n} - \paren {a_1 + a_2 + \cdots + a_m} }\)
\(\ds \) \(=\) \(\ds \cmod {a_{m + 1} + a_{m + 2} + \cdots + a_n}\) Indexed Summation over Adjacent Intervals
\(\ds \) \(\le\) \(\ds \cmod {a_{m + 1} } + \cmod {a_{m + 2} } + \cdots + \cmod {a_n}\) Triangle Inequality for Indexed Summations
\(\ds \) \(\le\) \(\ds b_{m + 1} + b_{m + 2} + \cdots + b_n\)
\(\ds \) \(\le\) \(\ds \sum_{k \mathop = n + 1}^\infty b_k\)
\(\ds \) \(<\) \(\ds \epsilon\)

So $\sequence {s_n}$ is a Cauchy sequence.

The result follows from Real Number Line is Complete Metric Space or Complex Plane is Complete Metric Space.

$\blacksquare$


Also see


Sources