determine series convergence.











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Determine whether this series converges:



$sum_{n=1}^infty cos(n^2pi) (sqrt{n+11} -sqrt{n+2}) $



I know that $lim a_{n} = 0$ and that this series alternates because of $cos(n^2pi)$, but don't know where to go from here.






sequences-and-series convergence conditional-convergence







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  • 1




    Squares of even and odd numbers are even and odd numbers, respectively
    – Mark Viola
    Nov 26 at 20:57















up vote
1
down vote

favorite












Determine whether this series converges:



$sum_{n=1}^infty cos(n^2pi) (sqrt{n+11} -sqrt{n+2}) $



I know that $lim a_{n} = 0$ and that this series alternates because of $cos(n^2pi)$, but don't know where to go from here.






sequences-and-series convergence conditional-convergence







share|cite|improve this question




















  • 1




    Squares of even and odd numbers are even and odd numbers, respectively
    – Mark Viola
    Nov 26 at 20:57













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Determine whether this series converges:



$sum_{n=1}^infty cos(n^2pi) (sqrt{n+11} -sqrt{n+2}) $



I know that $lim a_{n} = 0$ and that this series alternates because of $cos(n^2pi)$, but don't know where to go from here.






sequences-and-series convergence conditional-convergence







share|cite|improve this question















Determine whether this series converges:



$sum_{n=1}^infty cos(n^2pi) (sqrt{n+11} -sqrt{n+2}) $



I know that $lim a_{n} = 0$ and that this series alternates because of $cos(n^2pi)$, but don't know where to go from here.






sequences-and-series convergence conditional-convergence




sequences-and-series convergence conditional-convergence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 at 20:55









gimusi

90.6k74495




90.6k74495










asked Nov 26 at 20:49









J. Lastin

534




534








  • 1




    Squares of even and odd numbers are even and odd numbers, respectively
    – Mark Viola
    Nov 26 at 20:57














  • 1




    Squares of even and odd numbers are even and odd numbers, respectively
    – Mark Viola
    Nov 26 at 20:57








1




1




Squares of even and odd numbers are even and odd numbers, respectively
– Mark Viola
Nov 26 at 20:57




Squares of even and odd numbers are even and odd numbers, respectively
– Mark Viola
Nov 26 at 20:57










3 Answers
3






active

oldest

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up vote
4
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accepted










Observe that $$|a_n|=sqrt{n+11}-sqrt{n+2}=frac{n+11-n-2}{sqrt{n+11}+sqrt{n+2}}=frac{9}{sqrt{n+11}+sqrt{n+2}}$$ and that this sequence is decreasing and converges to zero. So, Leibniz criterion for alternate series says that it is convergent.






share|cite|improve this answer























  • I'm also supposed to determine whether the series converges conditionally or absolutely. Am I supposed to use the D'Alembert criterion for this?
    – J. Lastin
    Nov 26 at 21:39






  • 2




    The same calculation about $|a_n|$, says that, if you compare it with $1/sqrt{n}$, your series is NOT absolutely convergent.
    – Tito Eliatron
    Nov 26 at 21:41


















up vote
1
down vote













Try to use the alternating series test. Note that $a_{n} = sqrt{n+11} - sqrt{n+2}$ is a decreasing sequence (why?) and converges to 0. Also, we have $cos(n^{2}pi) = (-1)^{n}$.






share|cite|improve this answer




























    up vote
    1
    down vote













    Since the square of an even number is an even number and the square of an odd number is an odd number, then $cos(n^2pi)=(-1)^n$. Now appeal to Leibniz's Test.



    Can you finish?






    share|cite|improve this answer





















    • Good hint, even if I think that the asker was already aware about that! :)
      – gimusi
      Nov 26 at 21:38












    • @gimusi Thanks. It seemed that the OP had all of the conditions except monotonicity to apply Leibniz's Test
      – Mark Viola
      Nov 26 at 21:42










    • Yes indeed I think that the main doubt was on that! Cheers
      – gimusi
      Nov 26 at 21:43











    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    4
    down vote



    accepted










    Observe that $$|a_n|=sqrt{n+11}-sqrt{n+2}=frac{n+11-n-2}{sqrt{n+11}+sqrt{n+2}}=frac{9}{sqrt{n+11}+sqrt{n+2}}$$ and that this sequence is decreasing and converges to zero. So, Leibniz criterion for alternate series says that it is convergent.






    share|cite|improve this answer























    • I'm also supposed to determine whether the series converges conditionally or absolutely. Am I supposed to use the D'Alembert criterion for this?
      – J. Lastin
      Nov 26 at 21:39






    • 2




      The same calculation about $|a_n|$, says that, if you compare it with $1/sqrt{n}$, your series is NOT absolutely convergent.
      – Tito Eliatron
      Nov 26 at 21:41















    up vote
    4
    down vote



    accepted










    Observe that $$|a_n|=sqrt{n+11}-sqrt{n+2}=frac{n+11-n-2}{sqrt{n+11}+sqrt{n+2}}=frac{9}{sqrt{n+11}+sqrt{n+2}}$$ and that this sequence is decreasing and converges to zero. So, Leibniz criterion for alternate series says that it is convergent.






    share|cite|improve this answer























    • I'm also supposed to determine whether the series converges conditionally or absolutely. Am I supposed to use the D'Alembert criterion for this?
      – J. Lastin
      Nov 26 at 21:39






    • 2




      The same calculation about $|a_n|$, says that, if you compare it with $1/sqrt{n}$, your series is NOT absolutely convergent.
      – Tito Eliatron
      Nov 26 at 21:41













    up vote
    4
    down vote



    accepted







    up vote
    4
    down vote



    accepted






    Observe that $$|a_n|=sqrt{n+11}-sqrt{n+2}=frac{n+11-n-2}{sqrt{n+11}+sqrt{n+2}}=frac{9}{sqrt{n+11}+sqrt{n+2}}$$ and that this sequence is decreasing and converges to zero. So, Leibniz criterion for alternate series says that it is convergent.






    share|cite|improve this answer














    Observe that $$|a_n|=sqrt{n+11}-sqrt{n+2}=frac{n+11-n-2}{sqrt{n+11}+sqrt{n+2}}=frac{9}{sqrt{n+11}+sqrt{n+2}}$$ and that this sequence is decreasing and converges to zero. So, Leibniz criterion for alternate series says that it is convergent.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 27 at 1:43









    NoseKnowsAll

    1,8821020




    1,8821020










    answered Nov 26 at 20:54









    Tito Eliatron

    1,317622




    1,317622












    • I'm also supposed to determine whether the series converges conditionally or absolutely. Am I supposed to use the D'Alembert criterion for this?
      – J. Lastin
      Nov 26 at 21:39






    • 2




      The same calculation about $|a_n|$, says that, if you compare it with $1/sqrt{n}$, your series is NOT absolutely convergent.
      – Tito Eliatron
      Nov 26 at 21:41


















    • I'm also supposed to determine whether the series converges conditionally or absolutely. Am I supposed to use the D'Alembert criterion for this?
      – J. Lastin
      Nov 26 at 21:39






    • 2




      The same calculation about $|a_n|$, says that, if you compare it with $1/sqrt{n}$, your series is NOT absolutely convergent.
      – Tito Eliatron
      Nov 26 at 21:41
















    I'm also supposed to determine whether the series converges conditionally or absolutely. Am I supposed to use the D'Alembert criterion for this?
    – J. Lastin
    Nov 26 at 21:39




    I'm also supposed to determine whether the series converges conditionally or absolutely. Am I supposed to use the D'Alembert criterion for this?
    – J. Lastin
    Nov 26 at 21:39




    2




    2




    The same calculation about $|a_n|$, says that, if you compare it with $1/sqrt{n}$, your series is NOT absolutely convergent.
    – Tito Eliatron
    Nov 26 at 21:41




    The same calculation about $|a_n|$, says that, if you compare it with $1/sqrt{n}$, your series is NOT absolutely convergent.
    – Tito Eliatron
    Nov 26 at 21:41










    up vote
    1
    down vote













    Try to use the alternating series test. Note that $a_{n} = sqrt{n+11} - sqrt{n+2}$ is a decreasing sequence (why?) and converges to 0. Also, we have $cos(n^{2}pi) = (-1)^{n}$.






    share|cite|improve this answer

























      up vote
      1
      down vote













      Try to use the alternating series test. Note that $a_{n} = sqrt{n+11} - sqrt{n+2}$ is a decreasing sequence (why?) and converges to 0. Also, we have $cos(n^{2}pi) = (-1)^{n}$.






      share|cite|improve this answer























        up vote
        1
        down vote










        up vote
        1
        down vote









        Try to use the alternating series test. Note that $a_{n} = sqrt{n+11} - sqrt{n+2}$ is a decreasing sequence (why?) and converges to 0. Also, we have $cos(n^{2}pi) = (-1)^{n}$.






        share|cite|improve this answer












        Try to use the alternating series test. Note that $a_{n} = sqrt{n+11} - sqrt{n+2}$ is a decreasing sequence (why?) and converges to 0. Also, we have $cos(n^{2}pi) = (-1)^{n}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 26 at 20:52









        Seewoo Lee

        6,066826




        6,066826






















            up vote
            1
            down vote













            Since the square of an even number is an even number and the square of an odd number is an odd number, then $cos(n^2pi)=(-1)^n$. Now appeal to Leibniz's Test.



            Can you finish?






            share|cite|improve this answer





















            • Good hint, even if I think that the asker was already aware about that! :)
              – gimusi
              Nov 26 at 21:38












            • @gimusi Thanks. It seemed that the OP had all of the conditions except monotonicity to apply Leibniz's Test
              – Mark Viola
              Nov 26 at 21:42










            • Yes indeed I think that the main doubt was on that! Cheers
              – gimusi
              Nov 26 at 21:43















            up vote
            1
            down vote













            Since the square of an even number is an even number and the square of an odd number is an odd number, then $cos(n^2pi)=(-1)^n$. Now appeal to Leibniz's Test.



            Can you finish?






            share|cite|improve this answer





















            • Good hint, even if I think that the asker was already aware about that! :)
              – gimusi
              Nov 26 at 21:38












            • @gimusi Thanks. It seemed that the OP had all of the conditions except monotonicity to apply Leibniz's Test
              – Mark Viola
              Nov 26 at 21:42










            • Yes indeed I think that the main doubt was on that! Cheers
              – gimusi
              Nov 26 at 21:43













            up vote
            1
            down vote










            up vote
            1
            down vote









            Since the square of an even number is an even number and the square of an odd number is an odd number, then $cos(n^2pi)=(-1)^n$. Now appeal to Leibniz's Test.



            Can you finish?






            share|cite|improve this answer












            Since the square of an even number is an even number and the square of an odd number is an odd number, then $cos(n^2pi)=(-1)^n$. Now appeal to Leibniz's Test.



            Can you finish?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 26 at 21:00









            Mark Viola

            129k1273170




            129k1273170












            • Good hint, even if I think that the asker was already aware about that! :)
              – gimusi
              Nov 26 at 21:38












            • @gimusi Thanks. It seemed that the OP had all of the conditions except monotonicity to apply Leibniz's Test
              – Mark Viola
              Nov 26 at 21:42










            • Yes indeed I think that the main doubt was on that! Cheers
              – gimusi
              Nov 26 at 21:43


















            • Good hint, even if I think that the asker was already aware about that! :)
              – gimusi
              Nov 26 at 21:38












            • @gimusi Thanks. It seemed that the OP had all of the conditions except monotonicity to apply Leibniz's Test
              – Mark Viola
              Nov 26 at 21:42










            • Yes indeed I think that the main doubt was on that! Cheers
              – gimusi
              Nov 26 at 21:43
















            Good hint, even if I think that the asker was already aware about that! :)
            – gimusi
            Nov 26 at 21:38






            Good hint, even if I think that the asker was already aware about that! :)
            – gimusi
            Nov 26 at 21:38














            @gimusi Thanks. It seemed that the OP had all of the conditions except monotonicity to apply Leibniz's Test
            – Mark Viola
            Nov 26 at 21:42




            @gimusi Thanks. It seemed that the OP had all of the conditions except monotonicity to apply Leibniz's Test
            – Mark Viola
            Nov 26 at 21:42












            Yes indeed I think that the main doubt was on that! Cheers
            – gimusi
            Nov 26 at 21:43




            Yes indeed I think that the main doubt was on that! Cheers
            – gimusi
            Nov 26 at 21:43


















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