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Does $sum_n=2^ infty frac 1 n sqrt ln n$ converge?

0

$begingroup$

I want to figure out if this sum converges or diverges: $$sum_n=2^ infty frac 1 n sqrt ln n$$

I tried comparing it to the harmonic series, but this is less than that so it was no use. The limit comparison test with the harmonic series doesn't seem to work either, as it gives $infty$ or $0$. I thought of using the Integral Test, but this doesn't seem to have an obvious integral as far as I can tell.

How should this be done?

sequences-and-series divergent-series

share|cite|improve this question

edited Mar 19 at 11:37

Rócherz

3,0013821

asked Mar 15 '15 at 16:43

AskerAsker

806823

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Try the integral test, with the substitution $u=ln x$.
  $endgroup$
  – vadim123
  Mar 15 '15 at 16:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also see Bertrand's series.
  $endgroup$
  – Patissot
  Mar 15 '15 at 16:50
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  More generally, for any fixed $i$ the series $sum_ndfrac1nln nlnln nldots(ln^(i)n)^beta$ (where $ln^(i)$ denotes the $i$-times-iterated logarithm) converges iff $betagt 1$; this is the case $i=1$.
  $endgroup$
  – Steven Stadnicki
  Mar 15 '15 at 16:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The sum $sumlimits_n=2^inftyfrac1nln(n)$ does not converge, so your sum surely does not converge either.
  $endgroup$
  – barak manos
  Mar 15 '15 at 17:08
  

  
  
  
  

add a comment | 

0

$begingroup$

I want to figure out if this sum converges or diverges: $$sum_n=2^ infty frac 1 n sqrt ln n$$

I tried comparing it to the harmonic series, but this is less than that so it was no use. The limit comparison test with the harmonic series doesn't seem to work either, as it gives $infty$ or $0$. I thought of using the Integral Test, but this doesn't seem to have an obvious integral as far as I can tell.

How should this be done?

sequences-and-series divergent-series

share|cite|improve this question

edited Mar 19 at 11:37

Rócherz

3,0013821

asked Mar 15 '15 at 16:43

AskerAsker

806823

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Try the integral test, with the substitution $u=ln x$.
  $endgroup$
  – vadim123
  Mar 15 '15 at 16:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also see Bertrand's series.
  $endgroup$
  – Patissot
  Mar 15 '15 at 16:50
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  More generally, for any fixed $i$ the series $sum_ndfrac1nln nlnln nldots(ln^(i)n)^beta$ (where $ln^(i)$ denotes the $i$-times-iterated logarithm) converges iff $betagt 1$; this is the case $i=1$.
  $endgroup$
  – Steven Stadnicki
  Mar 15 '15 at 16:53
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The sum $sumlimits_n=2^inftyfrac1nln(n)$ does not converge, so your sum surely does not converge either.
  $endgroup$
  – barak manos
  Mar 15 '15 at 17:08
  

  
  
  
  

add a comment | 

$begingroup$

I want to figure out if this sum converges or diverges: $$sum_n=2^ infty frac 1 n sqrt ln n$$

I tried comparing it to the harmonic series, but this is less than that so it was no use. The limit comparison test with the harmonic series doesn't seem to work either, as it gives $infty$ or $0$. I thought of using the Integral Test, but this doesn't seem to have an obvious integral as far as I can tell.

How should this be done?

sequences-and-series divergent-series

share|cite|improve this question

edited Mar 19 at 11:37

Rócherz

3,0013821

asked Mar 15 '15 at 16:43

AskerAsker

806823

$endgroup$

share|cite|improve this question

edited Mar 19 at 11:37

Rócherz

3,0013821

asked Mar 15 '15 at 16:43

AskerAsker

806823

share|cite|improve this question

edited Mar 19 at 11:37

Rócherz

3,0013821

asked Mar 15 '15 at 16:43

AskerAsker

806823

edited Mar 19 at 11:37

Rócherz

3,0013821

edited Mar 19 at 11:37

Rócherz

3,0013821

asked Mar 15 '15 at 16:43

AskerAsker

806823

asked Mar 15 '15 at 16:43

AskerAsker

806823

3 Answers
3

active

oldest

votes

3

$begingroup$

By the Cauchy condensation test, your series is convergent iff
$$ sum_ngeq 2frac1sqrtn $$
is convergent, but obviously that is not the case.

share|cite|improve this answer

answered Mar 15 '15 at 16:53

Jack D'AurizioJack D'Aurizio

292k33284672

$endgroup$

add a comment | 

0

$begingroup$

With $u = ln(t)$:

$$int_2^inftyfrac1tsqrtln(t)dt = int_ln(2)^inftyfracdusqrtu = int_ln(2)^inftyfracduu^frac12$$

This diverges because the power is $frac12 le 1$.

By integral test, the series diverges.

share|cite|improve this answer

answered Mar 15 '15 at 16:49

user207710

$endgroup$

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  $begingroup$
  Why does it diverge because the power is $frac 1 2 le 1$? I'm guessing it has something to do with the rule for p-series, but how do we turn the integral back to a series?
  $endgroup$
  – Asker
  Mar 15 '15 at 17:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There's also an error in your notation; the $infty$ symbol on top of the last two integrals should actually be $ln (infty)$.
  $endgroup$
  – Asker
  Mar 15 '15 at 17:27
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Observe that $displaystyle sum_n=2^infty frac1n ln n= infty$, since $displaystyle fracddx[ln(ln(x))]=frac1x ln x$.

Notice that $displaystyle frac1nsqrtln n > frac1n ln n $, hence $displaystyle sum_n=2^infty frac1nsqrtln n= infty $.

share|cite|improve this answer

edited Mar 19 at 11:34

Rócherz

3,0013821

answered Mar 15 '15 at 16:55

studentstudent

706417

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think your $frac1sqrt n ln n$ should be $frac1nsqrt ln n$.
  $endgroup$
  – TonyK
  Mar 15 '15 at 16:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TonyK, thanks! I've edited the answer.
  $endgroup$
  – student
  Mar 15 '15 at 17:02
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  The same argument, more direct: $$fracddxleft(2sqrtln xright)=frac1xsqrtln x.$$
  $endgroup$
  – Did
  Mar 15 '15 at 17:14
  

  
  
  
  

add a comment | 

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3

$begingroup$

By the Cauchy condensation test, your series is convergent iff
$$ sum_ngeq 2frac1sqrtn $$
is convergent, but obviously that is not the case.

share|cite|improve this answer

answered Mar 15 '15 at 16:53

Jack D'AurizioJack D'Aurizio

292k33284672

$endgroup$

add a comment | 

3

$begingroup$

By the Cauchy condensation test, your series is convergent iff
$$ sum_ngeq 2frac1sqrtn $$
is convergent, but obviously that is not the case.

share|cite|improve this answer

answered Mar 15 '15 at 16:53

Jack D'AurizioJack D'Aurizio

292k33284672

$endgroup$

add a comment | 

$begingroup$

By the Cauchy condensation test, your series is convergent iff
$$ sum_ngeq 2frac1sqrtn $$
is convergent, but obviously that is not the case.

share|cite|improve this answer

answered Mar 15 '15 at 16:53

Jack D'AurizioJack D'Aurizio

292k33284672

$endgroup$

share|cite|improve this answer

answered Mar 15 '15 at 16:53

Jack D'AurizioJack D'Aurizio

292k33284672

answered Mar 15 '15 at 16:53

Jack D'AurizioJack D'Aurizio

292k33284672

answered Mar 15 '15 at 16:53

Jack D'AurizioJack D'Aurizio

292k33284672

0

$begingroup$

Observe that $displaystyle sum_n=2^infty frac1n ln n= infty$, since $displaystyle fracddx[ln(ln(x))]=frac1x ln x$.

Notice that $displaystyle frac1nsqrtln n > frac1n ln n $, hence $displaystyle sum_n=2^infty frac1nsqrtln n= infty $.

share|cite|improve this answer

edited Mar 19 at 11:34

Rócherz

3,0013821

answered Mar 15 '15 at 16:55

studentstudent

706417

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think your $frac1sqrt n ln n$ should be $frac1nsqrt ln n$.
  $endgroup$
  – TonyK
  Mar 15 '15 at 16:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TonyK, thanks! I've edited the answer.
  $endgroup$
  – student
  Mar 15 '15 at 17:02
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  The same argument, more direct: $$fracddxleft(2sqrtln xright)=frac1xsqrtln x.$$
  $endgroup$
  – Did
  Mar 15 '15 at 17:14
  

  
  
  
  

add a comment | 

0

$begingroup$

Observe that $displaystyle sum_n=2^infty frac1n ln n= infty$, since $displaystyle fracddx[ln(ln(x))]=frac1x ln x$.

Notice that $displaystyle frac1nsqrtln n > frac1n ln n $, hence $displaystyle sum_n=2^infty frac1nsqrtln n= infty $.

share|cite|improve this answer

edited Mar 19 at 11:34

Rócherz

3,0013821

answered Mar 15 '15 at 16:55

studentstudent

706417

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think your $frac1sqrt n ln n$ should be $frac1nsqrt ln n$.
  $endgroup$
  – TonyK
  Mar 15 '15 at 16:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @TonyK, thanks! I've edited the answer.
  $endgroup$
  – student
  Mar 15 '15 at 17:02
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  The same argument, more direct: $$fracddxleft(2sqrtln xright)=frac1xsqrtln x.$$
  $endgroup$
  – Did
  Mar 15 '15 at 17:14
  

  
  
  
  

add a comment | 

$begingroup$

Observe that $displaystyle sum_n=2^infty frac1n ln n= infty$, since $displaystyle fracddx[ln(ln(x))]=frac1x ln x$.

Notice that $displaystyle frac1nsqrtln n > frac1n ln n $, hence $displaystyle sum_n=2^infty frac1nsqrtln n= infty $.

share|cite|improve this answer

edited Mar 19 at 11:34

Rócherz

3,0013821

answered Mar 15 '15 at 16:55

studentstudent

706417

$endgroup$

share|cite|improve this answer

edited Mar 19 at 11:34

Rócherz

3,0013821

answered Mar 15 '15 at 16:55

studentstudent

706417

edited Mar 19 at 11:34

Rócherz

3,0013821

edited Mar 19 at 11:34

Rócherz

3,0013821

answered Mar 15 '15 at 16:55

studentstudent

706417

answered Mar 15 '15 at 16:55

studentstudent

706417