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CHAPTER 5 : SERIES - Coggle Diagram
CHAPTER 5 : SERIES
5.3 Test of convergence
Integral Test
Suppose ∞∑n=1
an
is a series with positive terms
an
. Suppose there exists a function f and a positive integer N such that the following three conditions are satisfied:
f is continuous,
f is decreasing, and
f(n)=an for all integers n≥N.
Ratio Test
L=limn→∞ (an+1/an)
Then, :
if L<1 the series is absolutely convergent (and hence convergent).
if L>1the series is divergent.
if L=1the series may be divergent, conditionally convergent, or absolutely convergent.
Divergence Test
the divergence test to determine whether a series converges or diverges.
If lim n→∞
an
= c≠0 or lim n→∞
an
does not exist, then the series ∑n=1∞an diverges.
For a series ∞∑n=1
an
to converge, the nth term
an
must satisfy
an
→0 as n→∞
5.5 Taylor and the Maclaurin Series
Finding Limits with Taylor Series and Maclaurin Series
if
f(x)
has a derivative of all orders at
x=a
, then we call the series as Taylor's series for
f(x)
about
x=a
and is given by
f(x) = f(a) + (x-a) f' (a) + (x-a)^2 / 2! f'' (a) + (x-a)^3 / 3! f''' (a) + ...
+ ((x-a)^r / r!) f^r (a)
In the special case where
a=0
, this series becomes the Maclaurin series for
f(x)
and is given by
f(x) = f(0) +x f'(0) + x^2/2! f'' (0) + ... + (x^r / r!) (f^r(0))
5.4 Power Series
Expansion of Logarithmic Function
ln (x) = ∞∑n=1 ((-1)n-1(x-1)n / n)
= (x-1) - (1/2)(x-1)^2 + (1/3)(x-1)^3 - (1/4)(x-1)^4 + ...
Taylor Series Centered at 1 (0 < x <=2)
-ln (x + 1) = ∞∑n=1 (x^n / n)
= x + (1/2)x^2 +(1/3)x^3 + (1/4)x^4 + ...
The series converges for (-1 < x <= 1)
Expansion of Trigonometric Function
sin x = x−x^3/3! +x^5/5! − x^7/7! +⋯
cos x = 1−x^2/2! + x^4/4! − x6^/6! +⋯
Expansion of Exponent Function
e^x = ∞∑n=0 (
xn
/ n!)
=1/1 + x/1 + x^2 / 2 + x^3 / 6 + ...
e = lim (n -> 0) (1 + n)^(1/n) or e = lim (n -> inf) (1 + 1/n)^n
2 The Sum of a Series
Sum of Power of ‘n’ Positive Integers
1st Power = n(n+1) / 2
2nd Power = (n/6)(n+1)(2n+1)
3rd Power = (n(n+1) / 2 )^2
Sum of Series of Partial Fraction
Terms with partial fractions such as : -
1/2.3 + 1/3.4 + 1/4.5 + ...
Sum of partial fractions requires a long solution but there is a different approach to solve the problem
5.1 Series
Sn
for the sum of these n terms
Consider the sequence of numbers
1, 2, 3, 4, 5, 6, . . . , n .
Then,
S1 = 1 = 1
S2 = 1+2 = 3
S3 = 1+2+3 = 6
The difference between the sum of two consecutive partial terms,
Sn - Sn-1
, is the nth term of the series.
Un = Sn - Sn-1
If the sum of the terms ends after a few terms, then he series is called
finite series.
If the sum of the series does not end, then the series is called
infinite series.
