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1 Sequences and Series
Definition: A sequence is a function whose domain is the set of integers: in other words for each integer we have a real number. Some examples:
Example: an = (^) n^12 or 1, 14 , 19 , 161...
Example: an = sin( nπ 4 ) 0,
√ 2 2 ,^1 ,^
√ 2 2 ,^0 ,^ −^
√ 2 2 ,^ −^1 ,...
Example: an = 2 + − n^1 1 , 52 , 53 , 94 , 95...
Definition: Convergence an → L if for every > 0 there exists an N such that |an − L| < for all n > N
In other words, the game is this: In order for a sequence to converge the following has to hold: You give me a tolerance () I have to find a number L (the limit) such that EVERY element after some point is within the tolerance () of the limit (L). If I can do this for EVERY tolerance then the sequence is convergent.
Examples: Examples 1 and 3 above are convergent. Example 2 is not convergent.
Proof an = 1 − (^) n^12. I claim that the limit is 1. In other words for every positive I can find N such that n > N implies |an − 0 | < . I can choose N = −^1 /^2
Of course one basically NEVER evaluates a limit using the , δ definition. The following theorem is the one which is most useful:
Theorem: Suppose that limx→∞ f (x) = L. Then limn→∞ f (n) = L
Example: Evaluate lim n→∞
2 n^2 + 1 3 n^2 + n + 18
lim x→∞
2 x^2 + 1 3 x^2 + x + 18
lim x→∞
4 x 6 x + 1
lim x→∞
BY the above theorem limx→∞ f (x) = limn→∞ f (n) = L, so we are done.
Example: Extended Estimate how large we need to take N so that |an − 2 3 | ≤^10
− 3 Note: The converse is not true. Take f (x) = sin(2π(x + (^) x^1 ))
Another useful result is the following:
Theorem: Suppose {an}∞ n=1 converges, and {bn}∞ n=1 also converges. Then
- limn→∞(an + bn) = lim an + lim bn
- limn→∞(an − bn) = lim an − lim bn
- limn→∞(anbn) = lim an × lim bn
- limn→∞(an/bn) = lim lim^ abnn if lim bn 6 = 0
- limn→∞ f (an) = f (limn→∞ an) if f is continuous
and
(Squeeze) Theorem: Suppose an, bn both converge to L, and there is an integer n 0 such that for n > n 0 we have an ≤ cn ≤ bn. The cn also converges to L.
Draw Picture!
Example: Consider the sequence an = n^ cos(n
(^2) )+e−n n^2.^ Note that^ −n^ ≤ n cos(n^2 ) ≤ n. Thus we have that (^1) n− 2 n ≤ an ≤ n n+1 2 Note that, by the L’Hopital’s rule (^1) n− 2 n → 0 and n n+1 2 → 0. Thus by the squeezing theorem an → 0
Corollary If |an| → 0 then an → 0 Pf: Observe that −|an| ≤ an ≤ |an|
Example an = (−1)
n n converges to 0
Flannery O’Connor Theorem: Everything that rises must converge
If a sequence is increasing and bounded above or decreasing and bounded below it converges.
Example: an = (^2) nn+1+5. Computing an+1 − an = (^) (n+1)(−^3 n+2) so an is decreas-
ing. an is bounded below. Thus an converges.
Example: an = n! n nn−^1 converges to zero.
A cautionary example: The above calculation does NOT by itself con- stitute a proof that the series converges. Consider the following example:
S =
∑^ ∞
n=
2 n^ = 2 + 4 + 8 + 16 + 32 +...
The same argument gives
S =
∑^ ∞
n=
2 n^ = 2 + 4 + 8 + 16 + 32 +...
2 S =
∑^ ∞
n=
2 n^ = 4 + 8 + 16 + 32 +...
−S = 2
Which is obviously Nonsense!. So actually summing the series is not enough - one must really check for convergence. However this method does suggest a method for proving convergence
SN =
∑^ N
n=
10 n^
10 N
SN =
10 N^
10 N^ +
So we have 9 10
SN =
10 N^ +
or
SN =
10 N
Since it is clear that (^101) N → 0 as N → ∞ we have SN → 19. If we do the cautionary example more carefully, we find something very different:
SN =
∑^ ∞
n=
2 n^ = 2 + 4 + 8 + 16 + 32 +... + 2N
2 SN =
∑^ ∞
n=
2 n^ = 4 + 8 + 16 + 32 +... + 2N^ + 2N^ +
−SN = 2 − 2 N^ +
SN = 2 N^ +1^ − 2
It is clear that the SN diverges! When we do the above manipulations we are assuming that the series converges. If this is NOT the case then the answer we get is meaningless.
So it is important to test for convergence. Let us recall some facts about geometric series: Example: Geometric Series
∑^ ∞
k=
ark
converges if |r| < 1. It diverges if |r| ≥ 1.
Example: Investigate the convergence of
∑^ ∞
k=
k(k + 2)
Note that
S 1 = 1 / 3 s 5 = 0. 595 S 100 =. 740 S 1000 =. 749 S 10000 =. 7499 S 1000000 = 0. 749999
