For which parameter this kind of series converges











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The sum is the following:
$sum_{k=1}^{infty}frac{a}{k^b}$



For which parameters this serie converges(a and b)?










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  • what have you tried? what result do you know that can possibly solve the problem?
    – Siong Thye Goh
    Nov 26 at 4:48















up vote
-2
down vote

favorite












The sum is the following:
$sum_{k=1}^{infty}frac{a}{k^b}$



For which parameters this serie converges(a and b)?










share|cite|improve this question






















  • what have you tried? what result do you know that can possibly solve the problem?
    – Siong Thye Goh
    Nov 26 at 4:48













up vote
-2
down vote

favorite









up vote
-2
down vote

favorite











The sum is the following:
$sum_{k=1}^{infty}frac{a}{k^b}$



For which parameters this serie converges(a and b)?










share|cite|improve this question













The sum is the following:
$sum_{k=1}^{infty}frac{a}{k^b}$



For which parameters this serie converges(a and b)?







convergence






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asked Nov 26 at 4:44









user587779

94




94












  • what have you tried? what result do you know that can possibly solve the problem?
    – Siong Thye Goh
    Nov 26 at 4:48


















  • what have you tried? what result do you know that can possibly solve the problem?
    – Siong Thye Goh
    Nov 26 at 4:48
















what have you tried? what result do you know that can possibly solve the problem?
– Siong Thye Goh
Nov 26 at 4:48




what have you tried? what result do you know that can possibly solve the problem?
– Siong Thye Goh
Nov 26 at 4:48










1 Answer
1






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0
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accepted










One may apply the integral test to $displaystyle f(x):=frac{a}{x^b}$,
$$
int_1^N f(x),dxlesum_{k=1}^N f(k)le f(1)+int_1^N f(x),dx,qquad N>1,
$$
and see it converges iff $,b>1$.






share|cite|improve this answer























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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote



    accepted










    One may apply the integral test to $displaystyle f(x):=frac{a}{x^b}$,
    $$
    int_1^N f(x),dxlesum_{k=1}^N f(k)le f(1)+int_1^N f(x),dx,qquad N>1,
    $$
    and see it converges iff $,b>1$.






    share|cite|improve this answer



























      up vote
      0
      down vote



      accepted










      One may apply the integral test to $displaystyle f(x):=frac{a}{x^b}$,
      $$
      int_1^N f(x),dxlesum_{k=1}^N f(k)le f(1)+int_1^N f(x),dx,qquad N>1,
      $$
      and see it converges iff $,b>1$.






      share|cite|improve this answer

























        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        One may apply the integral test to $displaystyle f(x):=frac{a}{x^b}$,
        $$
        int_1^N f(x),dxlesum_{k=1}^N f(k)le f(1)+int_1^N f(x),dx,qquad N>1,
        $$
        and see it converges iff $,b>1$.






        share|cite|improve this answer














        One may apply the integral test to $displaystyle f(x):=frac{a}{x^b}$,
        $$
        int_1^N f(x),dxlesum_{k=1}^N f(k)le f(1)+int_1^N f(x),dx,qquad N>1,
        $$
        and see it converges iff $,b>1$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 26 at 4:56

























        answered Nov 26 at 4:51









        Olivier Oloa

        107k17175293




        107k17175293






























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