Sequences CheatSheet

The two tests are perfectly equivalent; if one is inconclusive the other one is also inconclusive.
You can choose the most convenient one in a specific calculation. For example: determine for which
real numbers a the following sequence is convergent:
an = n!an
n n
When dealing with a factorial, the ratio test is always the best bet:
an+1
an
=
(n + 1)a
(n + 1)(1 + 1
n )n
The limit of the sequence is a
e . If a > e the sequence diverges, if a < e the sequence converges to
zero. If a = e the test is inconclusive. It can be shown by finer techniques that the case a = e is
actually divergent.
proof of prop. 4.7. If L < 1 we can choose ɛ > 0 such that L + ɛ < 1. According to the definition
of limit there exists nɛ ∈ N such that if n ≥ nɛ we have:
or equivalently:
L − ɛ < n |an| < L + ɛ < 1
0 ≤ |an| < (L + ɛ) n
Since L + ɛ < 1 the sequence (L + ɛ) n converges to zero and according to the squeeze theorem |an|
converges to zero as well. If the absolute value converges to zero, even an converges to zero.
If L > 1 we can choose ɛ > 0 such that L − ɛ > 1. If n ≥ nɛ we have:
(L − ɛ) n ≤ |an|
and since (L − ɛ) n diverges to +∞, the sequences |an| and an are also divergent.
proof of prop. 4.8. If L < 1 we can choose ɛ > 0 such that L + ɛ < 1. According to the definition
of limit there exists nɛ ∈ N such that if n ≥ nɛ we have:
< L + ɛ < 1
an+1
an
If n ≥ Nɛ we can apply the inequality repeatedly and obtain:
|an| ≤ (L + ɛ) n−nɛ anɛ
0 ≤ |an| < (L + ɛ) n
Since L + ɛ < 1 the sequence (L + ɛ) n−nɛ converges to zero and according to the squeeze theorem
|an| converges to zero as well. If the absolute value converges to zero, even an converges to zero.
If L > 1 we can choose ɛ > 0 such that L − ɛ > 1. If n ≥ nɛ we have:
(L − ɛ) ≤
Applying repeatedly we get:
an+1
an
(L − ɛ) n−nɛ anɛ ≤ |an|
and since (L − ɛ) n−nɛ− diverges to +∞, the sequences |an| and an are also divergent.
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