Question

What are two examples of convergent sequences?

Answer 1

Pick your favorites, or make your own!

Explanation:

Any constant sequence is convergent. For any `C in RR`
`lim_(n->oo)C = C`

• `lim_(n->oo)1 = 1`
• `lim_(n->oo)pi^e = pi^e`

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Any sequence in which the numerator is bounded and the denominator tends to `+-oo` will converge to `0`. That is, if `|a_n| < B` for some `B in RR` and `lim_(n->oo)b_n = oo` then
`lim_(n->oo)(a_n)/(b_n) = 0`

• `lim_(n->oo)1/n = 0`
• `lim_(n->oo)(100sin(n))/ln(n) = 0`

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If `|C| < 1` then `lim_(n->oo)C^n = 0`

• `lim_(n->oo)(1/2)^n = 0`
• `lim_(n->oo)(k/(k+1))^n = 0` for `k in [0, oo)`

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If `P(n)` and `Q(n)` are polynomials of order `k` where
`P(n) = a_0 + a_1n + a_2n^2 + ... + a_kn^k`
and
`Q(n) = b_0 + b_1n+b_2n^2 + ... + b_kn^k`

then
`lim_(n->oo)(P(n))/(Q(n)) = (a_k)/(b_k)`

• `lim_(n->oo)(2n+1)/(n+5) = 2/1 = 2`
• `lim_(n->oo)(4n^7-2n^2+1)/(-10n^7+10n^6+1000) = 4/(-10) = -2/5`

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1. `n^n`
2. `n!`
3. `C^n, |C| > 1`
4. `n^C`
5. `log(n)`

A term higher on the list divided by a term lower on the list will tend to `0`. This remains true if terms are added together, although not necessarily if they are multiplied. Multiplying by constants does not affect this.

• `lim_(n->oo) (n!)/n^n = 0`
• `lim_(n->oo) (2log(n) + n^2 + e^n)/(n!) = 0`

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If `|C| < 1` and `a in RR` then

`lim_(n->oo)sum_(k=0)^naC = a/(1-C)`

This is actually the geometric series formula.

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`lim_(n->oo)(1+1/n)^n = e`

In some places, this is how `e` is defined.

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There are many different ways to make convergent sequences. Some are intuitive. Some are not. Most require more justification than is provided here if the question is to show why they converge, but it is still useful to know what sorts of sequences converge and how.

And, finishing with a doozy, we have the Gaussian integral:

`lim_(n->oo)int_(-n)^n e^(-x^2)dx = sqrt(pi)`