infinite series containing cos(x/n)

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for $n in 1,2,3,dots $ we have $sumfrac(-1)^n+1cos(fracxn)n$. If integrated term by term twice, the resulting series clearly diverges. Does this mean the original series diverges as well? This is not a strictly alternating signs series and those have been difficult to prove the convergence/divergence of them.

sequences-and-series

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edited Jul 14 at 23:24

Mason

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asked Jul 14 at 22:32

D. J. R Miller

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  Here's a MathJax tutorial :)
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up vote
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for $n in 1,2,3,dots $ we have $sumfrac(-1)^n+1cos(fracxn)n$. If integrated term by term twice, the resulting series clearly diverges. Does this mean the original series diverges as well? This is not a strictly alternating signs series and those have been difficult to prove the convergence/divergence of them.

sequences-and-series

share|cite|improve this question

edited Jul 14 at 23:24

Mason

1,2401224

asked Jul 14 at 22:32

D. J. R Miller

1

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Here's a MathJax tutorial :)
  – Shaun
  Jul 14 at 22:35
  

  
  
  
  

add a comment | 

up vote
-4
down vote

favorite

up vote
-4
down vote

favorite

for $n in 1,2,3,dots $ we have $sumfrac(-1)^n+1cos(fracxn)n$. If integrated term by term twice, the resulting series clearly diverges. Does this mean the original series diverges as well? This is not a strictly alternating signs series and those have been difficult to prove the convergence/divergence of them.

sequences-and-series

share|cite|improve this question

edited Jul 14 at 23:24

Mason

1,2401224

asked Jul 14 at 22:32

D. J. R Miller

1

for $n in 1,2,3,dots $ we have $sumfrac(-1)^n+1cos(fracxn)n$. If integrated term by term twice, the resulting series clearly diverges. Does this mean the original series diverges as well? This is not a strictly alternating signs series and those have been difficult to prove the convergence/divergence of them.

sequences-and-series

share|cite|improve this question

edited Jul 14 at 23:24

Mason

1,2401224

asked Jul 14 at 22:32

D. J. R Miller

1

share|cite|improve this question

share|cite|improve this question

edited Jul 14 at 23:24

Mason

1,2401224

edited Jul 14 at 23:24

Mason

1,2401224

edited Jul 14 at 23:24

Mason

1,2401224

1,2401224

asked Jul 14 at 22:32

D. J. R Miller

1

asked Jul 14 at 22:32

D. J. R Miller

1

asked Jul 14 at 22:32

D. J. R Miller

1

1

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Here's a MathJax tutorial :)
  – Shaun
  Jul 14 at 22:35
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Here's a MathJax tutorial :)
  – Shaun
  Jul 14 at 22:35
  

  
  
  
  

2

2

Here's a MathJax tutorial :)
– Shaun
Jul 14 at 22:35

Here's a MathJax tutorial :)
– Shaun
Jul 14 at 22:35

add a comment | 

1 Answer
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For any specific $x$, it is eventually alternating. Also,$$frac ddnfraccos(x/n)n$$ is eventually negative, so the alternating series test applies, and it converges.

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answered Jul 14 at 22:45

Empy2

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1 Answer
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1 Answer
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For any specific $x$, it is eventually alternating. Also,$$frac ddnfraccos(x/n)n$$ is eventually negative, so the alternating series test applies, and it converges.

share|cite|improve this answer

answered Jul 14 at 22:45

Empy2

31.9k12059

add a comment | 

up vote
1
down vote

For any specific $x$, it is eventually alternating. Also,$$frac ddnfraccos(x/n)n$$ is eventually negative, so the alternating series test applies, and it converges.

share|cite|improve this answer

answered Jul 14 at 22:45

Empy2

31.9k12059

add a comment | 

up vote
1
down vote

up vote
1
down vote

For any specific $x$, it is eventually alternating. Also,$$frac ddnfraccos(x/n)n$$ is eventually negative, so the alternating series test applies, and it converges.

share|cite|improve this answer

answered Jul 14 at 22:45

Empy2

31.9k12059

For any specific $x$, it is eventually alternating. Also,$$frac ddnfraccos(x/n)n$$ is eventually negative, so the alternating series test applies, and it converges.

share|cite|improve this answer

answered Jul 14 at 22:45

Empy2

31.9k12059

share|cite|improve this answer

share|cite|improve this answer

answered Jul 14 at 22:45

Empy2

31.9k12059

answered Jul 14 at 22:45

Empy2

31.9k12059

answered Jul 14 at 22:45

Empy2

31.9k12059

31.9k12059

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