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

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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











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















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











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













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












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










share|cite|improve this question




share|cite|improve this question









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













  • 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











2 Answers
2






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




























    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




























      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

























        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























          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













          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













          share|cite|improve this answer



          share|cite|improve this answer











          answered Aug 3 at 20:25









          gimusi

          63.7k73480




          63.7k73480




















              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

























                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























                  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













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







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered Aug 3 at 20:26









                  J.G.

                  12.8k11323




                  12.8k11323












                      Comments

                      Post a Comment

                      Popular posts from this blog

                      What is the equation of a 3D cone with generalised tilt?

                      Image
                      Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
                      Read more

                      Relationship between determinant of matrix and determinant of adjoint?

                      Image
                      Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
                      Read more

                      Color the edges and diagonals of a regular polygon

                      Image
                      Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
                      Read more