The byproduct of sericulture in different industries.pptx
Review of series
1. MATH 1220 Summary of Convergence Tests for Series
∞
Let ∑a
n=1
n be an infinite series of positive terms.
∞
The series ∑a
n=1
n converges if and only if the sequence of partial sums,
∞
S n = a1 + a 2 + a3 + a n , converges. NOTE: lim S n = ∑a n
n→∞
n=1
∞
Divergence Test: If n → ∞ an ≠ 0 , the series
lim ∑a n diverges.
n=1
n 1
∞
n lim = lim =1
Example: The series ∑ is divergent since n→ ∞
n2 + 1 n→ ∞
1+ 1
n =1 n +12
n2
This means that the terms of a convergent series must approach zero. That is, if ∑ n
a
converges, then lim an = 0. However, lim a n = 0 does not imply convergence.
n →∞ n →∞
Geometric Series: THIS is our model series A geometric series
a + ar + ar 2 + + ar n −1 + converges for − 1 < r < 1 .
an +1 a
Note: r = If the series converges, the sum of the series is .
an 1 −r
n
∞
7 35 7
Example: The series ∑5 8 converges with a = a1 = 8 and r = 8 . The sum of the
n=1
series is 35.
Integral Test: If f is a continuous, positive, decreasing function on [1, ∞ with
)
∞
f ( n) = an , then the series ∑a
n=1
n converges if and only if the improper integral
∞
∫ f ( x)dx converges.
1
∞
Remainder for Integral Test: If ∑a
n=1
n converges by the Integral Test, then the
∞
remainder after n terms satisfies Rn ≤∫ f ( x)dx
n
∞
1
p-series: The series ∑n
n=1
p is convergent for p > 1 and diverges otherwise.
∞ ∞
1 1
Examples: The series ∑n
n=1
1.001 is convergent but the series ∑n
n=1
is divergent.
2. ∞ ∞
Comparison Test: Suppose ∑a
n=1
n and ∑b
n=1
n are series with positive terms.
∞ ∞
(a) If ∑b
n=1
n is convergent and an ≤ bn for all n, then
n=1
∑a n converges.
∞ ∞
(b) If ∑b
n=1
n is divergent and an ≥ bn for all n, then ∑a
n=1
n diverges.
The Comparison Test requires that you make one of two comparisons:
• Compare an unknown series to a LARGER known convergent series (smaller
than convergent is convergent)
• Compare an unknown series to a SMALLER known divergent series (bigger
than divergent is divergent)
∞ ∞ ∞
3n 3n 1
Examples: ∑n
n =2
2
> ∑ 2 = 3∑ which is a divergent harmonic series. Since the
− 2 n =2 n n =2 n
original series is larger by comparison, it is divergent.
∞ ∞
5n 5n 5 ∞ 1
We have ∑ 2n 3 + n 2 + 1 n=1 2n 2 n=1 n
n =1
< ∑ 3 = ∑ 2 which is a convergent p-series. Since the
original series is smaller by comparison, it is convergent.
∞ ∞
Limit Comparison Test: Suppose ∑a
n=1
n and ∑b
n=1
n are series with positive terms. If
an
lim = c where 0 < c < ∞ , then either both series converge or both series diverge.
n →∞ b
n
(Useful for p-series)
Rule of Thumb: To obtain a series for comparison, omit lower order terms in the
numerator and the denominator and then simplify.
∞ ∞
∞ n 1
Examples: For the series ∑ 2
n
n =1 n + n + 3
, compare to ∑n 2
=∑ 3 which is a
n =1 n =1 n 2
convergent p-series.
πn + n
n
∞ ∞
πn ∞
π
For the series ∑ n 2 , compare to ∑ n
= ∑ which is a divergent geometric
n =1 3 + n n =1 3 n =1 3
series.
Alternating Series Test: If the alternating series
∞
∑( −1)
n −1
bn = b1 − b2 + b3 − b4 + b5 − b6 +
n =1
satisfies (a) bn > bn +1 and (b) n →∞ bn = 0 , then the series converges.
lim
Remainder: Rn = s − sn ≤ bn +1
Absolute convergence simply means that the series converges without alternating (all
signs and terms are positive).
∞
( −1) n
Examples: The series ∑
n =0 n +1
is convergent but not absolutely convergent.
(− ) n
1 ∞
Alternating p-series: The alternating p-series ∑ p converges for p > 0.
n=1 n
3. ∞
( −1) n ∞
( −1) n
Examples: The series ∑
n=1 n
and the Alternating Harmonic series ∑
n=1 n
are
convergent.
4. ∞
an +1 an +1
lim
Ratio Test: (a) If n →∞
an
< 1 then the series ∑a
n=1
n lim
converges; (b) if n →∞
an
>1
the series diverges. Otherwise, you must use a different test for convergence.
This says that if the series eventually behaves like a convergent (divergent) geometric
series, it converges (diverges). If this limit is one, the test is inconclusive and a different
test is required. Specifically, the Ratio Test does not work for p-series.
POWER SERIES
∞
Radius of Convergence: The radius of convergence for a power series ∑c
n =0
n ( x − a) n
cn
is R = n →∞
lim . The center of the series is x = a. The series converges on the open
cn +1
interval ( a − R, a + R ) and may converge at the endpoints. You must test each series
that results at the endpoints of the interval separately for convergence.
∞
( x + 2) n
Example: The series ∑ (n +1)
n =0
2 is convergent on [-3,-1] but the series
∞
(−1) n ( x − 3) n
∑
n =0 5n n +1
is convergent on (-2,8].
Taylor Series: If f has a power series expansion centered at x = a, then the power series
∞
f ( n ) (a)
is given by f ( x) = ∑ ( x − a) n . Use the Ratio Test to determine the interval of
n =0 n!
convergence.
(n + )
Taylor’s Inequality (Remainder): If f
1
( x ) ≤ M for x − ≤d , then
a
M n +1
Rn ( x) ≤ x −a for x −a ≤d . Note that d < R (the radius of convergence)
( n +1)!
( n+ )
1
and think of M as the maximum value of f ( x ) on the interval [x,a] or [a,x].