Does $sum_k=1^infty frac 1 k^1+frac 1 k $ converge or diverge? [duplicate]

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This question already has an answer here:

• 
  Why doesn't $sum_n=1^infty frac1n^1+frac1n$ converge?
  
  2 answers
  
  

$sum_k=1^infty frac 1 k^1+frac 1 k $

What I've tried:

$a = $ $sum_k=1^infty frac 1 k^1+frac 1 k $

$b = $ $frac 1 k$

Limit comparison test:

$lim_nto infty frac a b = 1$

Therefore, by Limit comparison test, a and b diverge or converge together.

Because b diverges (p-series), a must also diverge.

Have I made any mistakes?

calculus sequences-and-series

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asked Aug 3 at 20:20

matthewninja

1424

marked as duplicate by Martin R, Community♦ Aug 3 at 20:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Also: math.stackexchange.com/q/1673312/42969, math.stackexchange.com/q/2054911/42969
  – Martin R
  Aug 3 at 20:24
  
  

  
  
  
  


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  For $kge1$ $$frac1k^1+1/kge frac1-frac1klog(k)k=frac1k - fraclog(k)k^2$$
  – Mark Viola
  Aug 3 at 20:34
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

This question already has an answer here:

• 
  Why doesn't $sum_n=1^infty frac1n^1+frac1n$ converge?
  
  2 answers
  
  

$sum_k=1^infty frac 1 k^1+frac 1 k $

What I've tried:

$a = $ $sum_k=1^infty frac 1 k^1+frac 1 k $

$b = $ $frac 1 k$

Limit comparison test:

$lim_nto infty frac a b = 1$

Therefore, by Limit comparison test, a and b diverge or converge together.

Because b diverges (p-series), a must also diverge.

Have I made any mistakes?

calculus sequences-and-series

share|cite|improve this question

asked Aug 3 at 20:20

matthewninja

1424

marked as duplicate by Martin R, Community♦ Aug 3 at 20:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Also: math.stackexchange.com/q/1673312/42969, math.stackexchange.com/q/2054911/42969
  – Martin R
  Aug 3 at 20:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For $kge1$ $$frac1k^1+1/kge frac1-frac1klog(k)k=frac1k - fraclog(k)k^2$$
  – Mark Viola
  Aug 3 at 20:34
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

This question already has an answer here:

• 
  Why doesn't $sum_n=1^infty frac1n^1+frac1n$ converge?
  
  2 answers
  
  

$sum_k=1^infty frac 1 k^1+frac 1 k $

What I've tried:

$a = $ $sum_k=1^infty frac 1 k^1+frac 1 k $

$b = $ $frac 1 k$

Limit comparison test:

$lim_nto infty frac a b = 1$

Therefore, by Limit comparison test, a and b diverge or converge together.

Because b diverges (p-series), a must also diverge.

Have I made any mistakes?

calculus sequences-and-series

share|cite|improve this question

asked Aug 3 at 20:20

matthewninja

1424

This question already has an answer here:

• 
  Why doesn't $sum_n=1^infty frac1n^1+frac1n$ converge?
  
  2 answers
  
  

$sum_k=1^infty frac 1 k^1+frac 1 k $

What I've tried:

$a = $ $sum_k=1^infty frac 1 k^1+frac 1 k $

$b = $ $frac 1 k$

Limit comparison test:

$lim_nto infty frac a b = 1$

Therefore, by Limit comparison test, a and b diverge or converge together.

Because b diverges (p-series), a must also diverge.

Have I made any mistakes?

This question already has an answer here:

• 
  Why doesn't $sum_n=1^infty frac1n^1+frac1n$ converge?
  
  2 answers
  
  

calculus sequences-and-series

share|cite|improve this question

asked Aug 3 at 20:20

matthewninja

1424

share|cite|improve this question

share|cite|improve this question

asked Aug 3 at 20:20

matthewninja

1424

asked Aug 3 at 20:20

matthewninja

1424

asked Aug 3 at 20:20

matthewninja

1424

1424

marked as duplicate by Martin R, Community♦ Aug 3 at 20:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Martin R, Community♦ Aug 3 at 20:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Also: math.stackexchange.com/q/1673312/42969, math.stackexchange.com/q/2054911/42969
  – Martin R
  Aug 3 at 20:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For $kge1$ $$frac1k^1+1/kge frac1-frac1klog(k)k=frac1k - fraclog(k)k^2$$
  – Mark Viola
  Aug 3 at 20:34
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Also: math.stackexchange.com/q/1673312/42969, math.stackexchange.com/q/2054911/42969
  – Martin R
  Aug 3 at 20:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For $kge1$ $$frac1k^1+1/kge frac1-frac1klog(k)k=frac1k - fraclog(k)k^2$$
  – Mark Viola
  Aug 3 at 20:34
  
  

  
  
  
  

1

1

Also: math.stackexchange.com/q/1673312/42969, math.stackexchange.com/q/2054911/42969
– Martin R
Aug 3 at 20:24

Also: math.stackexchange.com/q/1673312/42969, math.stackexchange.com/q/2054911/42969
– Martin R
Aug 3 at 20:24

For $kge1$ $$frac1k^1+1/kge frac1-frac1klog(k)k=frac1k - fraclog(k)k^2$$
– Mark Viola
Aug 3 at 20:34

For $kge1$ $$frac1k^1+1/kge frac1-frac1klog(k)k=frac1k - fraclog(k)k^2$$
– Mark Viola
Aug 3 at 20:34

add a comment | 

2 Answers
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accepted

Yes you are absolutely right indeed

$$fracfrac 1 k^1+frac 1 kfrac1k=frac k k^1+frac 1 k=frac 1 k^frac 1 kto 1$$

and by limit comparison test we can conclude.

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answered Aug 3 at 20:25

gimusi

63.7k73480

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Since $k^1/ksim 1$, $k^-1-1/ksim k^-1$ so the series diverges.

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answered Aug 3 at 20:26

J.G.

12.8k11323

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Yes you are absolutely right indeed

$$fracfrac 1 k^1+frac 1 kfrac1k=frac k k^1+frac 1 k=frac 1 k^frac 1 kto 1$$

and by limit comparison test we can conclude.

share|cite|improve this answer

answered Aug 3 at 20:25

gimusi

63.7k73480

add a comment | 

up vote
2
down vote



accepted

Yes you are absolutely right indeed

$$fracfrac 1 k^1+frac 1 kfrac1k=frac k k^1+frac 1 k=frac 1 k^frac 1 kto 1$$

and by limit comparison test we can conclude.

share|cite|improve this answer

answered Aug 3 at 20:25

gimusi

63.7k73480

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Yes you are absolutely right indeed

$$fracfrac 1 k^1+frac 1 kfrac1k=frac k k^1+frac 1 k=frac 1 k^frac 1 kto 1$$

and by limit comparison test we can conclude.

share|cite|improve this answer

answered Aug 3 at 20:25

gimusi

63.7k73480

Yes you are absolutely right indeed

$$fracfrac 1 k^1+frac 1 kfrac1k=frac k k^1+frac 1 k=frac 1 k^frac 1 kto 1$$

and by limit comparison test we can conclude.

share|cite|improve this answer

answered Aug 3 at 20:25

gimusi

63.7k73480

share|cite|improve this answer

share|cite|improve this answer

answered Aug 3 at 20:25

gimusi

63.7k73480

answered Aug 3 at 20:25

gimusi

63.7k73480

answered Aug 3 at 20:25

gimusi

63.7k73480

63.7k73480

add a comment | 

add a comment | 

up vote
2
down vote

Since $k^1/ksim 1$, $k^-1-1/ksim k^-1$ so the series diverges.

share|cite|improve this answer

answered Aug 3 at 20:26

J.G.

12.8k11323

add a comment | 

up vote
2
down vote

Since $k^1/ksim 1$, $k^-1-1/ksim k^-1$ so the series diverges.

share|cite|improve this answer

answered Aug 3 at 20:26

J.G.

12.8k11323

add a comment | 

up vote
2
down vote

up vote
2
down vote

Since $k^1/ksim 1$, $k^-1-1/ksim k^-1$ so the series diverges.

share|cite|improve this answer

answered Aug 3 at 20:26

J.G.

12.8k11323

Since $k^1/ksim 1$, $k^-1-1/ksim k^-1$ so the series diverges.

share|cite|improve this answer

answered Aug 3 at 20:26

J.G.

12.8k11323

share|cite|improve this answer

share|cite|improve this answer

answered Aug 3 at 20:26

J.G.

12.8k11323

answered Aug 3 at 20:26

J.G.

12.8k11323

answered Aug 3 at 20:26

J.G.

12.8k11323

12.8k11323

add a comment | 

add a comment |