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Page 1
Article D’Alembert
s Ratio Test
Statement If ?u
is a series of positive terms such that
(a) lim
u
u
then
(i) ?u
is convergent if (ii) ?u
is divergent if
(iii) ?u
may converge or diverge if (i e ,the test fails if )
(b) lim
u
u
,then the series ?u
is convergent
Proof (a) ?u
is series of positive terms
u
n
u
u
lim
u
u
Since lim
u
u
,therefore,for each a positive integer m such that
|
u
u
| n m
u
u
Replacing n by m,m ,m , ,n in the above inequality,we have
u
u
u
u
u
u
u
u
Multiplying the above (n m) inequality,we have
( )
u
u
( )
( )
(i) Let
Choose such that (This is always possible since we have only to choose an such that
From ( ),we have
( )
u
u
u
u
( )
u
u
( )
( )
u
K
( )
n m where K u
( )
Now,?
( )
being a geometric series with common ratio
is convergent Therefore,
Page 2
Article D’Alembert
s Ratio Test
Statement If ?u
is a series of positive terms such that
(a) lim
u
u
then
(i) ?u
is convergent if (ii) ?u
is divergent if
(iii) ?u
may converge or diverge if (i e ,the test fails if )
(b) lim
u
u
,then the series ?u
is convergent
Proof (a) ?u
is series of positive terms
u
n
u
u
lim
u
u
Since lim
u
u
,therefore,for each a positive integer m such that
|
u
u
| n m
u
u
Replacing n by m,m ,m , ,n in the above inequality,we have
u
u
u
u
u
u
u
u
Multiplying the above (n m) inequality,we have
( )
u
u
( )
( )
(i) Let
Choose such that (This is always possible since we have only to choose an such that
From ( ),we have
( )
u
u
u
u
( )
u
u
( )
( )
u
K
( )
n m where K u
( )
Now,?
( )
being a geometric series with common ratio
is convergent Therefore,
For more notes, call 8130648819
by comparison test,the series ? u
is convergent
(ii) Let
Choose such that (This is always possiblle since we have only to choose an such that
Since ,
From ( ),we have
u
u
( )
u
u
( )
u
u
( )
( )
u
( )
n m where u
( )
Now,?
( )
being a geometric series with common ratio
is divergent Therefore,
by comparison test,the series ? u
is divergent
(iii) Let
First consider the series
?
n
n
u
n
, u
n
u
u
n
n
n
so that
lim
u
u
Since the series ?
n
is divergent,we find that if ,a series may diverge
Next,consider the series
?
n
n
u
n
, u
(n )
u
u
(n )
n
(
n
n
)
(
n
)
so that Lim
u
u
Since the series ?
n
is convergent,we find that if
,a series may converge The above two examples show that if ,a series may converge or diverge
Hence the test fails when
Remar Another equivalent form of Ratio Test is as follow
If ?u
is a positive term series such that lim
u
u
then
(i) ?u
is convergent if
(ii) ?u
is divergent if
Ex Discuss the convergence of the following series
(i)
(ii)
,(p )
(iii)
(iv)
Solution (i) Here u
n
n
u
(n )
(n )
Page 3
Article D’Alembert
s Ratio Test
Statement If ?u
is a series of positive terms such that
(a) lim
u
u
then
(i) ?u
is convergent if (ii) ?u
is divergent if
(iii) ?u
may converge or diverge if (i e ,the test fails if )
(b) lim
u
u
,then the series ?u
is convergent
Proof (a) ?u
is series of positive terms
u
n
u
u
lim
u
u
Since lim
u
u
,therefore,for each a positive integer m such that
|
u
u
| n m
u
u
Replacing n by m,m ,m , ,n in the above inequality,we have
u
u
u
u
u
u
u
u
Multiplying the above (n m) inequality,we have
( )
u
u
( )
( )
(i) Let
Choose such that (This is always possible since we have only to choose an such that
From ( ),we have
( )
u
u
u
u
( )
u
u
( )
( )
u
K
( )
n m where K u
( )
Now,?
( )
being a geometric series with common ratio
is convergent Therefore,
For more notes, call 8130648819
by comparison test,the series ? u
is convergent
(ii) Let
Choose such that (This is always possiblle since we have only to choose an such that
Since ,
From ( ),we have
u
u
( )
u
u
( )
u
u
( )
( )
u
( )
n m where u
( )
Now,?
( )
being a geometric series with common ratio
is divergent Therefore,
by comparison test,the series ? u
is divergent
(iii) Let
First consider the series
?
n
n
u
n
, u
n
u
u
n
n
n
so that
lim
u
u
Since the series ?
n
is divergent,we find that if ,a series may diverge
Next,consider the series
?
n
n
u
n
, u
(n )
u
u
(n )
n
(
n
n
)
(
n
)
so that Lim
u
u
Since the series ?
n
is convergent,we find that if
,a series may converge The above two examples show that if ,a series may converge or diverge
Hence the test fails when
Remar Another equivalent form of Ratio Test is as follow
If ?u
is a positive term series such that lim
u
u
then
(i) ?u
is convergent if
(ii) ?u
is divergent if
Ex Discuss the convergence of the following series
(i)
(ii)
,(p )
(iii)
(iv)
Solution (i) Here u
n
n
u
(n )
(n )
For more notes, call 8130648819
u
u
n
n
(n )
(n )
n
n
(n )
(n )n
(n )
n
(
n
n
)
(
n
)
lim
u
u
lim
(
n
)
e , e -
y D’Alembert’s Ratio Test, u
is convergent
(ii) Do yourself ,Ans Convergent-
(iii) Do yourself ,Ans Convergent-
(iv) Do yourself ,Ans Convergent-
JAM MCQ Que
Let S be the series ?
( )
( )
and T be the series ? 4
5
( )
of real numbers
Then,which one of the following is true
(a) oth the series S and T are convergent
(b) S is convergent and T is divergent
(c) S is divergent and T is convergent
(d) oth the series S and T are divergent
Sol
S ?
( )
( )
?
convergent
or Consider the u
for S
u
( n )
( )
u
, (n ) -
, ( ) -
, n -,
-
u
u
( n )(
)
( n )(
)
n.
n
/(
)
n.
n
/(
)
.
n
/
.
n
/
lim
u
u
by ratio test S is convergent
Consider the u
for T
u
(
n
n
)
( )
u
(
n
n
)
( )
u
u
(
n
n
)
(
n
n
)
:
n
n
;
:
n
n
;
lim
u
u
tends to
So, T is divergent
Ex Test the convergence of the following series
(i)
(ii)
(iii)
(iv)
(v)
Solution (i) Here u
(n)
u
(n )
u
u
(n )
n
n
n
n
n
n
n
Page 4
Article D’Alembert
s Ratio Test
Statement If ?u
is a series of positive terms such that
(a) lim
u
u
then
(i) ?u
is convergent if (ii) ?u
is divergent if
(iii) ?u
may converge or diverge if (i e ,the test fails if )
(b) lim
u
u
,then the series ?u
is convergent
Proof (a) ?u
is series of positive terms
u
n
u
u
lim
u
u
Since lim
u
u
,therefore,for each a positive integer m such that
|
u
u
| n m
u
u
Replacing n by m,m ,m , ,n in the above inequality,we have
u
u
u
u
u
u
u
u
Multiplying the above (n m) inequality,we have
( )
u
u
( )
( )
(i) Let
Choose such that (This is always possible since we have only to choose an such that
From ( ),we have
( )
u
u
u
u
( )
u
u
( )
( )
u
K
( )
n m where K u
( )
Now,?
( )
being a geometric series with common ratio
is convergent Therefore,
For more notes, call 8130648819
by comparison test,the series ? u
is convergent
(ii) Let
Choose such that (This is always possiblle since we have only to choose an such that
Since ,
From ( ),we have
u
u
( )
u
u
( )
u
u
( )
( )
u
( )
n m where u
( )
Now,?
( )
being a geometric series with common ratio
is divergent Therefore,
by comparison test,the series ? u
is divergent
(iii) Let
First consider the series
?
n
n
u
n
, u
n
u
u
n
n
n
so that
lim
u
u
Since the series ?
n
is divergent,we find that if ,a series may diverge
Next,consider the series
?
n
n
u
n
, u
(n )
u
u
(n )
n
(
n
n
)
(
n
)
so that Lim
u
u
Since the series ?
n
is convergent,we find that if
,a series may converge The above two examples show that if ,a series may converge or diverge
Hence the test fails when
Remar Another equivalent form of Ratio Test is as follow
If ?u
is a positive term series such that lim
u
u
then
(i) ?u
is convergent if
(ii) ?u
is divergent if
Ex Discuss the convergence of the following series
(i)
(ii)
,(p )
(iii)
(iv)
Solution (i) Here u
n
n
u
(n )
(n )
For more notes, call 8130648819
u
u
n
n
(n )
(n )
n
n
(n )
(n )n
(n )
n
(
n
n
)
(
n
)
lim
u
u
lim
(
n
)
e , e -
y D’Alembert’s Ratio Test, u
is convergent
(ii) Do yourself ,Ans Convergent-
(iii) Do yourself ,Ans Convergent-
(iv) Do yourself ,Ans Convergent-
JAM MCQ Que
Let S be the series ?
( )
( )
and T be the series ? 4
5
( )
of real numbers
Then,which one of the following is true
(a) oth the series S and T are convergent
(b) S is convergent and T is divergent
(c) S is divergent and T is convergent
(d) oth the series S and T are divergent
Sol
S ?
( )
( )
?
convergent
or Consider the u
for S
u
( n )
( )
u
, (n ) -
, ( ) -
, n -,
-
u
u
( n )(
)
( n )(
)
n.
n
/(
)
n.
n
/(
)
.
n
/
.
n
/
lim
u
u
by ratio test S is convergent
Consider the u
for T
u
(
n
n
)
( )
u
(
n
n
)
( )
u
u
(
n
n
)
(
n
n
)
:
n
n
;
:
n
n
;
lim
u
u
tends to
So, T is divergent
Ex Test the convergence of the following series
(i)
(ii)
(iii)
(iv)
(v)
Solution (i) Here u
(n)
u
(n )
u
u
(n )
n
n
n
n
n
n
n
For more notes, call 8130648819
lim
u
u
lim
n
n
n
y D
Alembert
s Ratio Test,?u
is divergent
(ii) The given series is
Here u
n
u
(n )
u
u
n
(n )
n
n
.
n
/
.
n
/
lim
u
u
lim
.
n
/
y D
Alembert
s Ratio Test, u
s convergent
(iii) Do yourself ,Ans Divergent-
(iv) u
n
, u
(n )
divergent
u
u
n
(n )
n
lim
u
u
y D
Alembert Ratio Test,the series is divergent
(v) Do yourself ,Ans Divergent-
Ex Examine the convergence or divergence of the following series
(i)
(ii)
(iii) ?
Solution (i) Here
u
n
, (n ) -
n
( n )
u
n(n )
( n )( n )
u
u
n
( n )
( n )( n )
n(n )
u
u
( n )
n
n
n
lim
u
u
lim
n
n
y D
Alembert’s Ratio Test, u
is convergent
(ii) Here u
u
u
u
.
/
.
/
lim
u
u
lim
y D
Alembert’s Ratio Test, u
is divergent
(iii) Do yourself ,Ans Convergent-
Ex Test the convergence of the following series
(i) ?
n
(ii) ?
n
a
a
(iii) ?
n
(n )
n
(iv) ?
n
n
Solution (i) Do yourself ,Ans Convergent-
Page 5
Article D’Alembert
s Ratio Test
Statement If ?u
is a series of positive terms such that
(a) lim
u
u
then
(i) ?u
is convergent if (ii) ?u
is divergent if
(iii) ?u
may converge or diverge if (i e ,the test fails if )
(b) lim
u
u
,then the series ?u
is convergent
Proof (a) ?u
is series of positive terms
u
n
u
u
lim
u
u
Since lim
u
u
,therefore,for each a positive integer m such that
|
u
u
| n m
u
u
Replacing n by m,m ,m , ,n in the above inequality,we have
u
u
u
u
u
u
u
u
Multiplying the above (n m) inequality,we have
( )
u
u
( )
( )
(i) Let
Choose such that (This is always possible since we have only to choose an such that
From ( ),we have
( )
u
u
u
u
( )
u
u
( )
( )
u
K
( )
n m where K u
( )
Now,?
( )
being a geometric series with common ratio
is convergent Therefore,
For more notes, call 8130648819
by comparison test,the series ? u
is convergent
(ii) Let
Choose such that (This is always possiblle since we have only to choose an such that
Since ,
From ( ),we have
u
u
( )
u
u
( )
u
u
( )
( )
u
( )
n m where u
( )
Now,?
( )
being a geometric series with common ratio
is divergent Therefore,
by comparison test,the series ? u
is divergent
(iii) Let
First consider the series
?
n
n
u
n
, u
n
u
u
n
n
n
so that
lim
u
u
Since the series ?
n
is divergent,we find that if ,a series may diverge
Next,consider the series
?
n
n
u
n
, u
(n )
u
u
(n )
n
(
n
n
)
(
n
)
so that Lim
u
u
Since the series ?
n
is convergent,we find that if
,a series may converge The above two examples show that if ,a series may converge or diverge
Hence the test fails when
Remar Another equivalent form of Ratio Test is as follow
If ?u
is a positive term series such that lim
u
u
then
(i) ?u
is convergent if
(ii) ?u
is divergent if
Ex Discuss the convergence of the following series
(i)
(ii)
,(p )
(iii)
(iv)
Solution (i) Here u
n
n
u
(n )
(n )
For more notes, call 8130648819
u
u
n
n
(n )
(n )
n
n
(n )
(n )n
(n )
n
(
n
n
)
(
n
)
lim
u
u
lim
(
n
)
e , e -
y D’Alembert’s Ratio Test, u
is convergent
(ii) Do yourself ,Ans Convergent-
(iii) Do yourself ,Ans Convergent-
(iv) Do yourself ,Ans Convergent-
JAM MCQ Que
Let S be the series ?
( )
( )
and T be the series ? 4
5
( )
of real numbers
Then,which one of the following is true
(a) oth the series S and T are convergent
(b) S is convergent and T is divergent
(c) S is divergent and T is convergent
(d) oth the series S and T are divergent
Sol
S ?
( )
( )
?
convergent
or Consider the u
for S
u
( n )
( )
u
, (n ) -
, ( ) -
, n -,
-
u
u
( n )(
)
( n )(
)
n.
n
/(
)
n.
n
/(
)
.
n
/
.
n
/
lim
u
u
by ratio test S is convergent
Consider the u
for T
u
(
n
n
)
( )
u
(
n
n
)
( )
u
u
(
n
n
)
(
n
n
)
:
n
n
;
:
n
n
;
lim
u
u
tends to
So, T is divergent
Ex Test the convergence of the following series
(i)
(ii)
(iii)
(iv)
(v)
Solution (i) Here u
(n)
u
(n )
u
u
(n )
n
n
n
n
n
n
n
For more notes, call 8130648819
lim
u
u
lim
n
n
n
y D
Alembert
s Ratio Test,?u
is divergent
(ii) The given series is
Here u
n
u
(n )
u
u
n
(n )
n
n
.
n
/
.
n
/
lim
u
u
lim
.
n
/
y D
Alembert
s Ratio Test, u
s convergent
(iii) Do yourself ,Ans Divergent-
(iv) u
n
, u
(n )
divergent
u
u
n
(n )
n
lim
u
u
y D
Alembert Ratio Test,the series is divergent
(v) Do yourself ,Ans Divergent-
Ex Examine the convergence or divergence of the following series
(i)
(ii)
(iii) ?
Solution (i) Here
u
n
, (n ) -
n
( n )
u
n(n )
( n )( n )
u
u
n
( n )
( n )( n )
n(n )
u
u
( n )
n
n
n
lim
u
u
lim
n
n
y D
Alembert’s Ratio Test, u
is convergent
(ii) Here u
u
u
u
.
/
.
/
lim
u
u
lim
y D
Alembert’s Ratio Test, u
is divergent
(iii) Do yourself ,Ans Convergent-
Ex Test the convergence of the following series
(i) ?
n
(ii) ?
n
a
a
(iii) ?
n
(n )
n
(iv) ?
n
n
Solution (i) Do yourself ,Ans Convergent-
For more notes, call 8130648819
(ii) Here u
n
a
a
u
(n )
a
a
u
u
(n
a)(
a)
((n )
a)(
a)
n
.
a
n
/
n
6.
n
/
a
n
7
.
a
/
.
a
/
a
n
.
n
/
a
n
.
a
/
a
lim
u
u
y D
Alembert
s Ratio Test, u
is convergent
(iii) Do yourself ,Ans Convergent-
(iv) Do yourself ,Ans Convergent-
Ex Discuss the convergence or divergence of the following series
(i) ?
n
n
(ii) ?
n
n
(iii) ?
n
n
Solution (i) Here u
n
n
u
(n )
(n )
(n )
(n )n
n
n
u
u
n
n
n
n
lim
u
u
lim
n
n
y D
Alembert
s Ratio Test, u
is convergent
(ii) Do yourself ,Ans Convergent-
(iii) Do yourself ,Ans Divergent-
Ex Discuss the convergence or divergence of the following series
(i) ?
x
n
,x (ii) ?
x
a vn
(iii) ?
v
n
n
x
(iv) ?
vn
vn
x
(v) ?
x
n
,x (vi) ?
n
n
x
,x
Solution (i) Here u
x
n
u
x
(n )
u
u
x
x
(n )
n
x
(
n
)
lim
u
u
lim
x
(
n
)
x
y D
Alembert
s Ratio Test, u
converges if
x
i e x and diverges if
x
i e ,x
When x ,
lim
u
u
Ratio Test Fails
Now,when x ,
u
n
n
?u
?
n
is of the form ?
n
with p
?u
is convergent
Hence the given series u
converges if x and diverges if x
(ii) Do yourself ,Ans converges if x and diverges if x -
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