Courses

# Cauchy's Condensation Test (With Solved Exercise) Mathematics Notes | EduRev

## Mathematics : Cauchy's Condensation Test (With Solved Exercise) Mathematics Notes | EduRev

``` Page 2

For more notes,  call 8130648819

Since, logn is an increasing squence,
u

is a decreasing sequence
u

u

n
y Cauchy

s condensation test,the series ? u

and ?

u

converge or diverge together
Now
?

u

?

log

?

nlog

log
?

n

Consider v

n
v

n

lim

v

lim

n

y Cauchy

s root test,?v

is divergent
?

u

is divergent
? u

is divergent
(ii) Here
u

nlogn

Since, nlogn is an increasing sequence,
u

is a decreasing sequence
i e               u

u

n
y Cauchy

s condensation test,the series ? u

and ?

u

converge or diverge together
Now
?

u

?

log

?

nlog

log
?

n

Since,?

n

is divergent   ?

u

is divergent
? u

is divergent
(iii) Here
u

nvlogn

Since nvlogn is an increasing sequence,  u

is a decreasing sequence
u

u

n
y Cauchy

s condensation test,the series ? u

and ?

u

converge or diverge together
Now
?

u

?

vlog

?

vnlog

vlog
?

vn

Since,?

vn

is divergent                     ( p

)
?

u

is divergent
? u

is divergent
(iv) Do yourself
(v) Here u

n(logn)

Page 3

For more notes,  call 8130648819

Since, logn is an increasing squence,
u

is a decreasing sequence
u

u

n
y Cauchy

s condensation test,the series ? u

and ?

u

converge or diverge together
Now
?

u

?

log

?

nlog

log
?

n

Consider v

n
v

n

lim

v

lim

n

y Cauchy

s root test,?v

is divergent
?

u

is divergent
? u

is divergent
(ii) Here
u

nlogn

Since, nlogn is an increasing sequence,
u

is a decreasing sequence
i e               u

u

n
y Cauchy

s condensation test,the series ? u

and ?

u

converge or diverge together
Now
?

u

?

log

?

nlog

log
?

n

Since,?

n

is divergent   ?

u

is divergent
? u

is divergent
(iii) Here
u

nvlogn

Since nvlogn is an increasing sequence,  u

is a decreasing sequence
u

u

n
y Cauchy

s condensation test,the series ? u

and ?

u

converge or diverge together
Now
?

u

?

vlog

?

vnlog

vlog
?

vn

Since,?

vn

is divergent                     ( p

)
?

u

is divergent
? u

is divergent
(iv) Do yourself
(v) Here u

n(logn)

For more notes,  call 8130648819

Case I When p

n(logn)

n
n
Since,?

n

diverges,by comparison test ?

n(logn)

diverges
Case II When p
Since n(logn)

is an increasing sequence, u

is a decreasing sequence
i e                         u

u

n
y Cauchy

s condensation test,the series ? u

and ?

u

converge or diverge together
Now
?

u

?

(log

)

?

( log )

(log )

?

n

Since,?

n

is convergent if p   and divergent if p
?

u

is convergent if p   and divergent if p
? u

is convergent if p   and divergent if p
Hence,? u

is convergent if p   and divergent if p
(vi) Here u

(logn)

Case I When p  ,u

Since, lim

u

,?

diverges
Case II  when p  ,let p  q,where q
Since, lim

u

lim

(logn)

lim

(logn)

? u

diverges
Case III When  p
Since, (logn)

is an increasing sequence, u

is a decreasing sequence
i e                         u

u

n
y Cauchy

s  condensation  test,the series  ? u

and ?

u

converge or diverge together
Now  ?

u

?

(log

)

?

(nlog )

(log )

?

n

Consider, v

n

so that v

/

(n
/
)

lim

v

lim

(n

)

y Cauchy

s root test,?v

is divergent
?

u

is divergent
? u

is divergent
Hence,? u

is divergent for all values of p
Page 4

For more notes,  call 8130648819

Since, logn is an increasing squence,
u

is a decreasing sequence
u

u

n
y Cauchy

s condensation test,the series ? u

and ?

u

converge or diverge together
Now
?

u

?

log

?

nlog

log
?

n

Consider v

n
v

n

lim

v

lim

n

y Cauchy

s root test,?v

is divergent
?

u

is divergent
? u

is divergent
(ii) Here
u

nlogn

Since, nlogn is an increasing sequence,
u

is a decreasing sequence
i e               u

u

n
y Cauchy

s condensation test,the series ? u

and ?

u

converge or diverge together
Now
?

u

?

log

?

nlog

log
?

n

Since,?

n

is divergent   ?

u

is divergent
? u

is divergent
(iii) Here
u

nvlogn

Since nvlogn is an increasing sequence,  u

is a decreasing sequence
u

u

n
y Cauchy

s condensation test,the series ? u

and ?

u

converge or diverge together
Now
?

u

?

vlog

?

vnlog

vlog
?

vn

Since,?

vn

is divergent                     ( p

)
?

u

is divergent
? u

is divergent
(iv) Do yourself
(v) Here u

n(logn)

For more notes,  call 8130648819

Case I When p

n(logn)

n
n
Since,?

n

diverges,by comparison test ?

n(logn)

diverges
Case II When p
Since n(logn)

is an increasing sequence, u

is a decreasing sequence
i e                         u

u

n
y Cauchy

s condensation test,the series ? u

and ?

u

converge or diverge together
Now
?

u

?

(log

)

?

( log )

(log )

?

n

Since,?

n

is convergent if p   and divergent if p
?

u

is convergent if p   and divergent if p
? u

is convergent if p   and divergent if p
Hence,? u

is convergent if p   and divergent if p
(vi) Here u

(logn)

Case I When p  ,u

Since, lim

u

,?

diverges
Case II  when p  ,let p  q,where q
Since, lim

u

lim

(logn)

lim

(logn)

? u

diverges
Case III When  p
Since, (logn)

is an increasing sequence, u

is a decreasing sequence
i e                         u

u

n
y Cauchy

s  condensation  test,the series  ? u

and ?

u

converge or diverge together
Now  ?

u

?

(log

)

?

(nlog )

(log )

?

n

Consider, v

n

so that v

/

(n
/
)

lim

v

lim

(n

)

y Cauchy

s root test,?v

is divergent
?

u

is divergent
? u

is divergent
Hence,? u

is divergent for all values of p
For more notes,  call 8130648819

Ex    Using Cauchy

s condensation test,discuss the convergence of ?
logn
n

Solution Here
u

logn
n
n
Consider         f(x)
logx
x
,x
f

(x)
x

x
logx
x

logx
x

f

(x)
logx
logx
x e                                ( loge  )
f(x) is a decreasing function when x e
u

u

(   e  )
u

is a decreasing sequence of positive terms
y Cauchy

s condensation test,the series ? u

and ?

u

converge or diverge together
Now,                ?

u

?

log

? nlog

log ? n

Since,? n

is divergent,?

u

is divergent
? u

is divergent
Ex    Using Cauchy

s condensation test,discuss the convergence of ? (
logn
n
)

Solution Here
u

(
logn
n
)

n
consider         f(x) (
logx
x
)

,                                 x
f

(x)  (
logx
x
) (
logx
x

)
f

(x)                   logx
logx          x e    ( loge  )
f(x) is decreasing function where x e
u

u

n                               (   e  )
u

is a decreasing sequence of positive terms
y Cauchy

s condensation test,the series ? u

and ?

u

converge or diverge together
Now
?

u

?

4
log

5

?

(nlog )

(log )

?
n

Consider v

n

so that v

y Cauchy

s root test,? v

converges
?

u

converges ? u

converges
Page 5

For more notes,  call 8130648819

Since, logn is an increasing squence,
u

is a decreasing sequence
u

u

n
y Cauchy

s condensation test,the series ? u

and ?

u

converge or diverge together
Now
?

u

?

log

?

nlog

log
?

n

Consider v

n
v

n

lim

v

lim

n

y Cauchy

s root test,?v

is divergent
?

u

is divergent
? u

is divergent
(ii) Here
u

nlogn

Since, nlogn is an increasing sequence,
u

is a decreasing sequence
i e               u

u

n
y Cauchy

s condensation test,the series ? u

and ?

u

converge or diverge together
Now
?

u

?

log

?

nlog

log
?

n

Since,?

n

is divergent   ?

u

is divergent
? u

is divergent
(iii) Here
u

nvlogn

Since nvlogn is an increasing sequence,  u

is a decreasing sequence
u

u

n
y Cauchy

s condensation test,the series ? u

and ?

u

converge or diverge together
Now
?

u

?

vlog

?

vnlog

vlog
?

vn

Since,?

vn

is divergent                     ( p

)
?

u

is divergent
? u

is divergent
(iv) Do yourself
(v) Here u

n(logn)

For more notes,  call 8130648819

Case I When p

n(logn)

n
n
Since,?

n

diverges,by comparison test ?

n(logn)

diverges
Case II When p
Since n(logn)

is an increasing sequence, u

is a decreasing sequence
i e                         u

u

n
y Cauchy

s condensation test,the series ? u

and ?

u

converge or diverge together
Now
?

u

?

(log

)

?

( log )

(log )

?

n

Since,?

n

is convergent if p   and divergent if p
?

u

is convergent if p   and divergent if p
? u

is convergent if p   and divergent if p
Hence,? u

is convergent if p   and divergent if p
(vi) Here u

(logn)

Case I When p  ,u

Since, lim

u

,?

diverges
Case II  when p  ,let p  q,where q
Since, lim

u

lim

(logn)

lim

(logn)

? u

diverges
Case III When  p
Since, (logn)

is an increasing sequence, u

is a decreasing sequence
i e                         u

u

n
y Cauchy

s  condensation  test,the series  ? u

and ?

u

converge or diverge together
Now  ?

u

?

(log

)

?

(nlog )

(log )

?

n

Consider, v

n

so that v

/

(n
/
)

lim

v

lim

(n

)

y Cauchy

s root test,?v

is divergent
?

u

is divergent
? u

is divergent
Hence,? u

is divergent for all values of p
For more notes,  call 8130648819

Ex    Using Cauchy

s condensation test,discuss the convergence of ?
logn
n

Solution Here
u

logn
n
n
Consider         f(x)
logx
x
,x
f

(x)
x

x
logx
x

logx
x

f

(x)
logx
logx
x e                                ( loge  )
f(x) is a decreasing function when x e
u

u

(   e  )
u

is a decreasing sequence of positive terms
y Cauchy

s condensation test,the series ? u

and ?

u

converge or diverge together
Now,                ?

u

?

log

? nlog

log ? n

Since,? n

is divergent,?

u

is divergent
? u

is divergent
Ex    Using Cauchy

s condensation test,discuss the convergence of ? (
logn
n
)

Solution Here
u

(
logn
n
)

n
consider         f(x) (
logx
x
)

,                                 x
f

(x)  (
logx
x
) (
logx
x

)
f

(x)                   logx
logx          x e    ( loge  )
f(x) is decreasing function where x e
u

u

n                               (   e  )
u

is a decreasing sequence of positive terms
y Cauchy

s condensation test,the series ? u

and ?

u

converge or diverge together
Now
?

u

?

4
log

5

?

(nlog )

(log )

?
n

Consider v

n

so that v

y Cauchy

s root test,? v

converges
?

u

converges ? u

converges
For more notes,  call 8130648819

Ex     Using Cauchy

s  condensation test,discuss the convergence of  ?

(nlogn)

Solution  Here  u

(nlogn)

,       n
Case I  When  p  ,u

Since, lim

u

, ? u

is divergent
Case II When p  ,let p  q  where q
u

(nlogn)

(nlogn)

u

as n
Since, lim

u

, u

is divergent
Case III When p  , u

is a decreasing sequence
y Cauchy

s condensation test,the series  ? u

and ?

u

converge or diverge together
Now       ?

u

?

(

log

)

?

(

nlog )

?

(   )
n

(log )

(log )

?

(   )
n

If p  ,then
(   )
so that

(   )
and

(   )
n

n

ut  the series ?

n

is convergent for p
?

(   )
n

is convegent   ?

u

is convergent for  p
If  p  ,then  ?

u

log
?

n

ut ?

n
is divergent
?

u

is divergent for p
If p  ,then    p    so that
(   )
,i e

(   )
and

(   )
and

(   )
n

n

ut  the series ?

n

is divergent for p
Hence,? u

is convergent for p   and divergent for p
Def A series with terms alternatively positive and negative is called an alternating series
Thus the series
u

u

u

u

where u

n
is an alternating series and is briefly written as ?(  )

u

Leibnitz

s Test on Alternating Series
Statement   The alternating series
?(  )

u

u

u

u

u

(u

n) converges if
(i) u

u

n and (ii) lim

u

Proof Let S

denote the nth partial sum of the series   ?(  )

u

S

u

u

u

u

u

u

u

u

u

(u

u

) (u

u

)  (u

u

) u

u

,(u

u

) (u

u

)   (u

u

) u

-
u

,   u

u

and u

n-
the sequence S

is bounded above
Also      S

S

u

u

```
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

## Topic-wise Tests & Solved Examples for IIT JAM Mathematics

27 docs|150 tests

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;