| Arithmetic Sequences |
Geometric Sequences |
A. Sequence
represents an ordered list of numbers with a common difference d.
Ex.: 1, 5, 9, 13, 17, 21, 25, ... ,
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G. Sequence
represents an ordered list of numbers with a common ratio r.
Ex.: 1, 4, 16, 64, 256, 1024, ... ,
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A. Series and G. Series
are the values obtain by adding up all the terms of a sequence. there also called the "sum".
Ex.: 1 + 5 + 9 + 13 + 17 + 21 + 25 + ... + (ASs or APs)
Ex.: 1 + 4 + 16 + 64 + 256 + 1024 + ... + (GSs or GPs)
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Common difference
is always the same value from one term to the next by adding (or subtracting) both terms.
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Ex: d = 9 − 5 = 4 (ASs above)
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Common ratio
is always the same value from one term to the next by dividing (or multiplying) both terms.
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the nth number
an = a1 + (n − 1)·d
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the nth number
an = a1 · rn − 1
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Sum of the nth 1st numbers
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Sum of the nth 1st numbers
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Sn = |
a1(rn − 1)
r − 1
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= |
a(1 − rn)
1 − r
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| with a1 = a, and the special circumstance that the r is between –1 and 1, | r | < 1. |
| the infinite sum has the following value: |
| S∞ = |
a
1 − r
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, with | r | < 1 |
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Sum of a Finite Geometric Series
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Let's find the sum of the first 7 terms of sequence S, with a1 = 1, r = 2 and n = 7.
We would then plug those numbers into the formula and get: |
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S7 = |
1(1 − 27)
1 − 2
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= |
(1 − 128)
− 1
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= |
− 127
− 1
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= |
127 |
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Sum of an infinite Geometric Series
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An infinite number of terms a sequence S in given by: 16 + (-8) + 4 + (-2) + 1 + ...,
with a1 = 16 and r = −1/2. r lies within the acceptable limits (− 1 and 1), therefore a finite sum exists when using the infinite term sum formula:
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| S∞ = |
a1
1 − r
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= |
16
1 − (−1/2)
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= |
16
3/2
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= 16 |
2
3
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= |
10.67 |
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| Convergence and Divergence of Sequences
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Examples
- Showing convergence or divergence of the the following sequence:
1, 2, 3, 4, ...
Ans. This sequence starts with and goes on bigger and bigger to the infinity (∞), therefore the sequence Diverges.
- Does the following sequence converges or diverges?
1, 0.1, 0.01, 0.001, 0.0001, ...
Ans. The sequence gets smaller and smaller and closer to 0, therefore it converges.
- Assuming we have the nth number of a sequence define by:
Ans. To show weither the sequence converges or diverges we have 2 possibilities:
- we can first find the 3 or 4 first terms of the sequence
| - a2 = |
3(−1)2
2!
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= |
3
2·1
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= |
3
2
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| - a3 = |
3(−1)3
3!
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= |
−3
3·2·1
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= |
−1
2
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As you can see, the terms are getting smaller and smaller and tending to 0, hence the sequence converges, (not obvious though!!!)
That's why we will use limits to make it clear...
lim
n→∞ |
[ |
3(−1)n
n!
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] = |
∞
∞
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(indeterminate form!) |
we use L'Hôpital's rule (by derivating several times until we get the form allowing the determination of the limits)
the second differentiation gives us:
lim
n→∞ |
[ |
3n(−1)n−1
1!
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] = |
∞·0
1!
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= 0 |
Now, we can clearly see that the sequence converges, as it equals to 0
- an = n(n −1) converges or diverges?
lim n(n − 1) = ∞·∞ = ∞ ⇨ the sequence diverges!
n→∞
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