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Ms. Deepali Gupta. JIIT, NOIDA
Convergence of Series
A series is an expression of the form u1 + u2 + u3 + ..+ un in
which the successive terms follow some regular law.
For e.g.
1 + 3 + 5 + 7+.+ 2n-1+
1+ x + x2 + x3 + ..+ xn + ., |x| < 1
1 + 2x + 3x2 + 4x3 +
If a series terminates at some particular term, it is called a finite
series.
If the number of terms is unlimited, it is called an infinite series.
++++ ..........8
1
4
1
2
11
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Ms. Deepali Gupta. JIIT, NOIDA
Let Sn
be sum of first n terms of the infinite series un
i.e. Sn
= u1
+ u2+ u3 +.. + un. Then the infinite series is said to be convergent to a
number S if Sn tends to a finite limit S as n increases indefinitely.
SSlim nn=
The series Sn is said to be divergent if
=
orSlim nn
The series Sn is said to be oscillatory if it does not tend to a unique
limit, finite or infinite.t
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Ms. Deepali Gupta. JIIT, NOIDA
Example
Sn
= 1+ 2 + 3 + + n
= n(n+1)/2
+
=
2
)1n(nlimSlimn
nn
Hence, this series is divergent.
Sn = 2 2 + 2 2 + 2 .
S1 = 2, S2 = 0, S3 = 2, S4 = 0
.yoscillatorisitHence.existnotdoesSlimnn
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Ms. Deepali Gupta. JIIT, NOIDA
Geometric Series
=
=+++++1n
1n1n2 ar........ar.....arara
1n2
n ar.....araraS++++=
n1n2
n arar.....ararSr ++++=
)r1(aS)r1( nn =
r1
ar
r1
a
)r1(
)r1(aSlim
nn
nn
=
=
1|r|ICase < 0rthan n
SeriesConvergentr1
a
Slim nn =
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Ms. Deepali Gupta. JIIT, NOIDA
1|r|IICase > nrthan
seriesDivergentSlim nn
=
Case III r = 1
a + a + a + ..+ a = na Divergent Series
r < -1, let r = -k where k > 1nnn
k)1(r =
))k(1(
)k)1(1(a
)r1(
)r1(aSlim
nnn
nn
=
=
evenisnif-oddisnif
== Oscillatory Series
r > 1
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Ms. Deepali Gupta. JIIT, NOIDA
++ .........8421 Oscillatory Series
++++ .........4
27
2
932 Divergent Series
5 5 5 .........4 16 64 + +
Convergent Series to -1
Case IV r = -1
a - a + a - ..
Sn= a, if n is odd
Sn= 0, if n is even Oscillatory Series
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Ms. Deepali Gupta. JIIT, NOIDA
N t h Term Test
0ulimthenconverges,uIfnn1n n
=
=
not true.isaboveofconverseThediverge.mayIt.convergentbenotmayseriesthen the0ulimife.i n
n
=
e.convergencforconditionsufficientanotbutconditionnecessaryaisulimHence nn
n21n u..........uuS:Pf +++=n1nn uSS +=
S.beinfinityuptosumitsand,convergentbeseriestheLet
SSlim nn
=
SSlim 1nn
=
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8/3/2019 Convergence Lec. 1
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Ms. Deepali Gupta. JIIT, NOIDA
++++++ ..........
n
1.......
4
1
3
1
2
11
n
1.......
4
1
3
1
2
11Sn +++++=
n1u n = 0
n1ulim n
n==
n
1...........
n
1
n
1
n
1.......
4
1
3
1
2
11 ++++++++
nn
n==
=
n
n
n
Slim
nS
Hence the series diverges.
8/3/2019 Convergence Lec. 1
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Ms. Deepali Gupta. JIIT, NOIDA
Rat io Test
Let un be a series of positive terms such that thenruulim
n
1n =+
(a) The series converges if r < 1.
(b) The series diverges if r > 1.
(c) The test fails for r = 1.
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Ms. Deepali Gupta. JIIT, NOIDA
Example
Test the series ...........x
4
5x
3
4x
2
32 32 ++++
1n
n xn
)1n(u
+=
n
1n x)1n(
)2n(u
+
+=+
x)1n(
)2n(n
u
ulim
2
n
1n
+
+=+ x=
Hence if x < 1, the series is convergent; and if x > 1, the series is divergent.
If x = 1 ...........4
5
3
4
2
32 ++++
01n
1nlimu n =+
= Thus the series is divergent.
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Ms. Deepali Gupta. JIIT, NOIDA
Root Test
Let un be a series of positive terms such that then
(a) The series converges if r < 1.
(b) The series diverges if r > 1.
(c) The test fails for r = 1.
nn
nlim u r
=
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Ms. Deepali Gupta. JIIT, NOIDA
Example
Discuss the convergence of the series
+
+
+
.............3
4
3
4
2
3
2
3
1
2
1
23
4
42
3
31
2
2
n
1n
1n
nn
1n
n
)1n(u
+
+
++= ( )1
1n
1nn/1
nn
1n
n
)1n(u
+
+
++=
( )
11n
n
n/1n
n n11
n11limulim
+
+
+=
1n
n n11
n11
n11lim
+
+
+=
( ) 11e1
1e1
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Ms. Deepali Gupta. JIIT, NOIDA
Cauc hy s In t egra l Test
A positive term series f(1) + f(2) + f(3) ++ f(n) +. where f(n)
decreases as n increases, converges or diverges according to the integral
infinite.orfiniteisdx)x(f1
=
1n
n2ne
2xxe)x(f =
Examine the convergence of
=
1
x
m1
x
2
elimdxxe
2
2
+
2
e
2
elim
1m
m
2
finite.ise2
1=
Hence the series is convergent by Integral Test .
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Ms. Deepali Gupta. JIIT, NOIDA
Show that the p-series
.........n1..........
31
21
11
n1
pppp1n
p+++++=
=
( p a real constant ) converges if p > 1, and diverges if p 1
x.offunctiondecreasingpositiveaisx1f(x)then1,pIf
p=>
dxxdx
x
1
1
p
1 p
=m
1
1p
m
1p
xlim
+
=+
=
1
m
1lim
p1
11pm
finite.is1p
1
=
The series converges by Cauchys Integral Test for p > 1.
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Ms. Deepali Gupta. JIIT, NOIDA
When p < 1
=
dx)x(f1
( )
1mlim
p1
1 p1m
The series diverges by Cauchys Integral Test for p < 1.
When p = 1
.........
n
1..........
3
1
2
1
1
1+++++
[ ] ==
111 xlogdxx1
dx)x(f
The series diverges by Cauchys Integral Test for p = 1.
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