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Comparison Test

Suppose that converges absolutely, and is a sequence of numbers for which | bn | | an | for all n > N. Then the series converges absolutely as well.

If the series converges to positive infinity, and is a sequence of numbers for which an bn for all n > N. Then the series also diverges.

This is a useful test, but the limit comparison test, which is rather similar, is a much easier to use, and therefore more useful. However, this comparison test is very easy to memorize: Assuming that everything is positive, for simplicity, say we know that:
| b n | | a n |
for all n. then just sum both sides to see what you get formally:
    • Then:
    Examples 4.2.4:
     
    • Does converge or diverge ?
    • Does converge or diverge ?
    Proof:

    The proof, at first glance, seems easy: Suppose that converges absolutely, and | b n | | a n | for all n. For simplicity, assume that all terms in both sequences are positive. Let

    S N = and T N =
    Then we have that
    T N SN
    Since the left side is a convergent sequence, it is in particular bounded. Hence, the right side is also a bounded sequence of partial sums. Therefore it converges.

    This proof wrong, because it does show that the sequence of partial sums is bounded but it is not necessarily true that a bounded series also converges - as we know.

    However, this proof, slightly modified, does work: Again, assume that all terms in both sequences are positive. Since converges, it satisfies the Cauchy criterion:

    | | <
    if m > n > N. Since | b n | | a n | we then have
    | | | | <
    if m > n > N. Hence satisfies the Cauchy criterion, and therefore converges.

    The proof for divergence is similar.

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