Showing $sum_{k=0}^{n} {n choose k}frac{{(-1)}^k}{n+k+1}$ is positive












4












$begingroup$



Show that the sum$$sum_{k=0}^{n} {n choose k}frac{{(-1)}^k}{n+k+1}$$ is a positive rational number.




It is easy to show that it is a rational number. But I am having trouble showing that this expression is positive. It might be some binomial expansion that I could not get.










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  • 1




    $begingroup$
    Have you tried using induction on $n$ for example?
    $endgroup$
    – Minus One-Twelfth
    yesterday


















4












$begingroup$



Show that the sum$$sum_{k=0}^{n} {n choose k}frac{{(-1)}^k}{n+k+1}$$ is a positive rational number.




It is easy to show that it is a rational number. But I am having trouble showing that this expression is positive. It might be some binomial expansion that I could not get.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Have you tried using induction on $n$ for example?
    $endgroup$
    – Minus One-Twelfth
    yesterday
















4












4








4


4



$begingroup$



Show that the sum$$sum_{k=0}^{n} {n choose k}frac{{(-1)}^k}{n+k+1}$$ is a positive rational number.




It is easy to show that it is a rational number. But I am having trouble showing that this expression is positive. It might be some binomial expansion that I could not get.










share|cite|improve this question











$endgroup$





Show that the sum$$sum_{k=0}^{n} {n choose k}frac{{(-1)}^k}{n+k+1}$$ is a positive rational number.




It is easy to show that it is a rational number. But I am having trouble showing that this expression is positive. It might be some binomial expansion that I could not get.







combinatorics summation binomial-coefficients binomial-ideals






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edited yesterday









YuiTo Cheng

2,1192837




2,1192837










asked yesterday









Hitendra KumarHitendra Kumar

706




706








  • 1




    $begingroup$
    Have you tried using induction on $n$ for example?
    $endgroup$
    – Minus One-Twelfth
    yesterday
















  • 1




    $begingroup$
    Have you tried using induction on $n$ for example?
    $endgroup$
    – Minus One-Twelfth
    yesterday










1




1




$begingroup$
Have you tried using induction on $n$ for example?
$endgroup$
– Minus One-Twelfth
yesterday






$begingroup$
Have you tried using induction on $n$ for example?
$endgroup$
– Minus One-Twelfth
yesterday












3 Answers
3






active

oldest

votes


















15












$begingroup$

Direct proof:
$$begin{split}
sum_{k=0}^{n} {nchoose k}frac{{(-1)}^k}{n+k+1} &=sum_{k=0}^{n} {nchoose k}(-1)^kint_0^1 x^{n+k}dx\
&=int_0^1x^nsum_{k=0}^{n} {nchoose k}(-x)^kdx\
&=int_0^1x^n(1-x)^ndx
end{split}$$

The latter is clearly a positive number.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How did you conclude that the sum is a limited integral? Do you know where can I find more on this on-line? Thanks.
    $endgroup$
    – NoChance
    yesterday






  • 4




    $begingroup$
    It's a "known trick" that $frac 1 {p+1} = int_0^1x^pdx$. Then I noticed that the sum looked almost like that of the binomial theorem.
    $endgroup$
    – Stefan Lafon
    yesterday



















7












$begingroup$

When $k=0$ the term is positive. When $k=1$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=0$ TERM.



When $k=2$ the term is positive. When $k=3$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=2$ TERM.



.....



Get it?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    sorry, I did not write question correctly. Now, I have corrected that. By looking at your answer I realized my mistake. Thanks
    $endgroup$
    – Hitendra Kumar
    yesterday



















6












$begingroup$

We can specifically prove that
$$
boxed{sum_{k=0}^{n} {n choose k}frac{{(-1)}^k}{n+k+1}=left((2n+1)binom{2n}nright)^{-1}}
$$

To see this, shuffle a deck of $2n+1$ cards numbered $1$ to $2n+1$. Consider this:




What is the probability that card number $n+1$ is in the middle of the deck, and cards numbered $1$ to $n$ are below it?




The easy answer is the fraction on the RHS. The LHS can be interpreted as an application of the principle of inclusion exclusion. Namely, we first take the probability that card number of $n+1$ is the lowest of the cards numbered $n+1,n+2,dots,2n+1$. This is the $k=0$ term. From this, for each $i=1,dots,n$, we subtract the probability that $n+1$ is the lowest of the list $i,n+1,n+2,dots,2n+1$. This is a bad event, because we want $n+1$ to be above $i$. Doing this for each $i$, we subtract $binom{n}1frac{1}{n+2}$. We then must add back in the doubly subtracted events, subtract the triple intersections, and so on, eventually ending with the alternating sum on the left.






share|cite|improve this answer











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    protected by Community yesterday



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    Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).



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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    15












    $begingroup$

    Direct proof:
    $$begin{split}
    sum_{k=0}^{n} {nchoose k}frac{{(-1)}^k}{n+k+1} &=sum_{k=0}^{n} {nchoose k}(-1)^kint_0^1 x^{n+k}dx\
    &=int_0^1x^nsum_{k=0}^{n} {nchoose k}(-x)^kdx\
    &=int_0^1x^n(1-x)^ndx
    end{split}$$

    The latter is clearly a positive number.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      How did you conclude that the sum is a limited integral? Do you know where can I find more on this on-line? Thanks.
      $endgroup$
      – NoChance
      yesterday






    • 4




      $begingroup$
      It's a "known trick" that $frac 1 {p+1} = int_0^1x^pdx$. Then I noticed that the sum looked almost like that of the binomial theorem.
      $endgroup$
      – Stefan Lafon
      yesterday
















    15












    $begingroup$

    Direct proof:
    $$begin{split}
    sum_{k=0}^{n} {nchoose k}frac{{(-1)}^k}{n+k+1} &=sum_{k=0}^{n} {nchoose k}(-1)^kint_0^1 x^{n+k}dx\
    &=int_0^1x^nsum_{k=0}^{n} {nchoose k}(-x)^kdx\
    &=int_0^1x^n(1-x)^ndx
    end{split}$$

    The latter is clearly a positive number.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      How did you conclude that the sum is a limited integral? Do you know where can I find more on this on-line? Thanks.
      $endgroup$
      – NoChance
      yesterday






    • 4




      $begingroup$
      It's a "known trick" that $frac 1 {p+1} = int_0^1x^pdx$. Then I noticed that the sum looked almost like that of the binomial theorem.
      $endgroup$
      – Stefan Lafon
      yesterday














    15












    15








    15





    $begingroup$

    Direct proof:
    $$begin{split}
    sum_{k=0}^{n} {nchoose k}frac{{(-1)}^k}{n+k+1} &=sum_{k=0}^{n} {nchoose k}(-1)^kint_0^1 x^{n+k}dx\
    &=int_0^1x^nsum_{k=0}^{n} {nchoose k}(-x)^kdx\
    &=int_0^1x^n(1-x)^ndx
    end{split}$$

    The latter is clearly a positive number.






    share|cite|improve this answer









    $endgroup$



    Direct proof:
    $$begin{split}
    sum_{k=0}^{n} {nchoose k}frac{{(-1)}^k}{n+k+1} &=sum_{k=0}^{n} {nchoose k}(-1)^kint_0^1 x^{n+k}dx\
    &=int_0^1x^nsum_{k=0}^{n} {nchoose k}(-x)^kdx\
    &=int_0^1x^n(1-x)^ndx
    end{split}$$

    The latter is clearly a positive number.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered yesterday









    Stefan LafonStefan Lafon

    3,005212




    3,005212












    • $begingroup$
      How did you conclude that the sum is a limited integral? Do you know where can I find more on this on-line? Thanks.
      $endgroup$
      – NoChance
      yesterday






    • 4




      $begingroup$
      It's a "known trick" that $frac 1 {p+1} = int_0^1x^pdx$. Then I noticed that the sum looked almost like that of the binomial theorem.
      $endgroup$
      – Stefan Lafon
      yesterday


















    • $begingroup$
      How did you conclude that the sum is a limited integral? Do you know where can I find more on this on-line? Thanks.
      $endgroup$
      – NoChance
      yesterday






    • 4




      $begingroup$
      It's a "known trick" that $frac 1 {p+1} = int_0^1x^pdx$. Then I noticed that the sum looked almost like that of the binomial theorem.
      $endgroup$
      – Stefan Lafon
      yesterday
















    $begingroup$
    How did you conclude that the sum is a limited integral? Do you know where can I find more on this on-line? Thanks.
    $endgroup$
    – NoChance
    yesterday




    $begingroup$
    How did you conclude that the sum is a limited integral? Do you know where can I find more on this on-line? Thanks.
    $endgroup$
    – NoChance
    yesterday




    4




    4




    $begingroup$
    It's a "known trick" that $frac 1 {p+1} = int_0^1x^pdx$. Then I noticed that the sum looked almost like that of the binomial theorem.
    $endgroup$
    – Stefan Lafon
    yesterday




    $begingroup$
    It's a "known trick" that $frac 1 {p+1} = int_0^1x^pdx$. Then I noticed that the sum looked almost like that of the binomial theorem.
    $endgroup$
    – Stefan Lafon
    yesterday











    7












    $begingroup$

    When $k=0$ the term is positive. When $k=1$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=0$ TERM.



    When $k=2$ the term is positive. When $k=3$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=2$ TERM.



    .....



    Get it?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      sorry, I did not write question correctly. Now, I have corrected that. By looking at your answer I realized my mistake. Thanks
      $endgroup$
      – Hitendra Kumar
      yesterday
















    7












    $begingroup$

    When $k=0$ the term is positive. When $k=1$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=0$ TERM.



    When $k=2$ the term is positive. When $k=3$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=2$ TERM.



    .....



    Get it?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      sorry, I did not write question correctly. Now, I have corrected that. By looking at your answer I realized my mistake. Thanks
      $endgroup$
      – Hitendra Kumar
      yesterday














    7












    7








    7





    $begingroup$

    When $k=0$ the term is positive. When $k=1$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=0$ TERM.



    When $k=2$ the term is positive. When $k=3$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=2$ TERM.



    .....



    Get it?






    share|cite|improve this answer









    $endgroup$



    When $k=0$ the term is positive. When $k=1$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=0$ TERM.



    When $k=2$ the term is positive. When $k=3$ the term is negative BUT SMALLER (in absolute value) THAN THE $k=2$ TERM.



    .....



    Get it?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered yesterday









    David G. StorkDavid G. Stork

    11.2k41432




    11.2k41432












    • $begingroup$
      sorry, I did not write question correctly. Now, I have corrected that. By looking at your answer I realized my mistake. Thanks
      $endgroup$
      – Hitendra Kumar
      yesterday


















    • $begingroup$
      sorry, I did not write question correctly. Now, I have corrected that. By looking at your answer I realized my mistake. Thanks
      $endgroup$
      – Hitendra Kumar
      yesterday
















    $begingroup$
    sorry, I did not write question correctly. Now, I have corrected that. By looking at your answer I realized my mistake. Thanks
    $endgroup$
    – Hitendra Kumar
    yesterday




    $begingroup$
    sorry, I did not write question correctly. Now, I have corrected that. By looking at your answer I realized my mistake. Thanks
    $endgroup$
    – Hitendra Kumar
    yesterday











    6












    $begingroup$

    We can specifically prove that
    $$
    boxed{sum_{k=0}^{n} {n choose k}frac{{(-1)}^k}{n+k+1}=left((2n+1)binom{2n}nright)^{-1}}
    $$

    To see this, shuffle a deck of $2n+1$ cards numbered $1$ to $2n+1$. Consider this:




    What is the probability that card number $n+1$ is in the middle of the deck, and cards numbered $1$ to $n$ are below it?




    The easy answer is the fraction on the RHS. The LHS can be interpreted as an application of the principle of inclusion exclusion. Namely, we first take the probability that card number of $n+1$ is the lowest of the cards numbered $n+1,n+2,dots,2n+1$. This is the $k=0$ term. From this, for each $i=1,dots,n$, we subtract the probability that $n+1$ is the lowest of the list $i,n+1,n+2,dots,2n+1$. This is a bad event, because we want $n+1$ to be above $i$. Doing this for each $i$, we subtract $binom{n}1frac{1}{n+2}$. We then must add back in the doubly subtracted events, subtract the triple intersections, and so on, eventually ending with the alternating sum on the left.






    share|cite|improve this answer











    $endgroup$


















      6












      $begingroup$

      We can specifically prove that
      $$
      boxed{sum_{k=0}^{n} {n choose k}frac{{(-1)}^k}{n+k+1}=left((2n+1)binom{2n}nright)^{-1}}
      $$

      To see this, shuffle a deck of $2n+1$ cards numbered $1$ to $2n+1$. Consider this:




      What is the probability that card number $n+1$ is in the middle of the deck, and cards numbered $1$ to $n$ are below it?




      The easy answer is the fraction on the RHS. The LHS can be interpreted as an application of the principle of inclusion exclusion. Namely, we first take the probability that card number of $n+1$ is the lowest of the cards numbered $n+1,n+2,dots,2n+1$. This is the $k=0$ term. From this, for each $i=1,dots,n$, we subtract the probability that $n+1$ is the lowest of the list $i,n+1,n+2,dots,2n+1$. This is a bad event, because we want $n+1$ to be above $i$. Doing this for each $i$, we subtract $binom{n}1frac{1}{n+2}$. We then must add back in the doubly subtracted events, subtract the triple intersections, and so on, eventually ending with the alternating sum on the left.






      share|cite|improve this answer











      $endgroup$
















        6












        6








        6





        $begingroup$

        We can specifically prove that
        $$
        boxed{sum_{k=0}^{n} {n choose k}frac{{(-1)}^k}{n+k+1}=left((2n+1)binom{2n}nright)^{-1}}
        $$

        To see this, shuffle a deck of $2n+1$ cards numbered $1$ to $2n+1$. Consider this:




        What is the probability that card number $n+1$ is in the middle of the deck, and cards numbered $1$ to $n$ are below it?




        The easy answer is the fraction on the RHS. The LHS can be interpreted as an application of the principle of inclusion exclusion. Namely, we first take the probability that card number of $n+1$ is the lowest of the cards numbered $n+1,n+2,dots,2n+1$. This is the $k=0$ term. From this, for each $i=1,dots,n$, we subtract the probability that $n+1$ is the lowest of the list $i,n+1,n+2,dots,2n+1$. This is a bad event, because we want $n+1$ to be above $i$. Doing this for each $i$, we subtract $binom{n}1frac{1}{n+2}$. We then must add back in the doubly subtracted events, subtract the triple intersections, and so on, eventually ending with the alternating sum on the left.






        share|cite|improve this answer











        $endgroup$



        We can specifically prove that
        $$
        boxed{sum_{k=0}^{n} {n choose k}frac{{(-1)}^k}{n+k+1}=left((2n+1)binom{2n}nright)^{-1}}
        $$

        To see this, shuffle a deck of $2n+1$ cards numbered $1$ to $2n+1$. Consider this:




        What is the probability that card number $n+1$ is in the middle of the deck, and cards numbered $1$ to $n$ are below it?




        The easy answer is the fraction on the RHS. The LHS can be interpreted as an application of the principle of inclusion exclusion. Namely, we first take the probability that card number of $n+1$ is the lowest of the cards numbered $n+1,n+2,dots,2n+1$. This is the $k=0$ term. From this, for each $i=1,dots,n$, we subtract the probability that $n+1$ is the lowest of the list $i,n+1,n+2,dots,2n+1$. This is a bad event, because we want $n+1$ to be above $i$. Doing this for each $i$, we subtract $binom{n}1frac{1}{n+2}$. We then must add back in the doubly subtracted events, subtract the triple intersections, and so on, eventually ending with the alternating sum on the left.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited yesterday

























        answered yesterday









        Mike EarnestMike Earnest

        25.6k22151




        25.6k22151

















            protected by Community yesterday



            Thank you for your interest in this question.
            Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).



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