Showing a sum is positive












2












$begingroup$



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.






combinatorics summation binomial-coefficients binomial-ideals







share|cite|improve this question











$endgroup$








  • 1




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


















2












$begingroup$



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.






combinatorics summation binomial-coefficients binomial-ideals







share|cite|improve this question











$endgroup$








  • 1




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
















2












2








2


1



$begingroup$



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.






combinatorics summation binomial-coefficients binomial-ideals







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




combinatorics summation binomial-coefficients binomial-ideals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago







Hitendra Kumar

















asked 1 hour ago









Hitendra KumarHitendra Kumar

606




606








  • 1




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
















  • 1




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










1




1




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






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












3 Answers
3






active

oldest

votes


















4












$begingroup$

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









$endgroup$













  • $begingroup$
    Thanks,I got it.
    $endgroup$
    – Hitendra Kumar
    48 mins ago










  • $begingroup$
    You're welcome!
    $endgroup$
    – Stefan Lafon
    47 mins ago










  • $begingroup$
    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
    47 mins ago






  • 1




    $begingroup$
    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
    41 mins ago










  • $begingroup$
    Thanks for responding.Got it.
    $endgroup$
    – NoChance
    31 mins ago





















3












$begingroup$

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
    1 hour ago



















0












$begingroup$

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 $n$. 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









$endgroup$













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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    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









    $endgroup$













    • $begingroup$
      Thanks,I got it.
      $endgroup$
      – Hitendra Kumar
      48 mins ago










    • $begingroup$
      You're welcome!
      $endgroup$
      – Stefan Lafon
      47 mins ago










    • $begingroup$
      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
      47 mins ago






    • 1




      $begingroup$
      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
      41 mins ago










    • $begingroup$
      Thanks for responding.Got it.
      $endgroup$
      – NoChance
      31 mins ago


















    4












    $begingroup$

    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









    $endgroup$













    • $begingroup$
      Thanks,I got it.
      $endgroup$
      – Hitendra Kumar
      48 mins ago










    • $begingroup$
      You're welcome!
      $endgroup$
      – Stefan Lafon
      47 mins ago










    • $begingroup$
      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
      47 mins ago






    • 1




      $begingroup$
      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
      41 mins ago










    • $begingroup$
      Thanks for responding.Got it.
      $endgroup$
      – NoChance
      31 mins ago
















    4












    4








    4





    $begingroup$

    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 1 hour ago









    Stefan LafonStefan Lafon

    3,00019




    3,00019












    • $begingroup$
      Thanks,I got it.
      $endgroup$
      – Hitendra Kumar
      48 mins ago










    • $begingroup$
      You're welcome!
      $endgroup$
      – Stefan Lafon
      47 mins ago










    • $begingroup$
      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
      47 mins ago






    • 1




      $begingroup$
      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
      41 mins ago










    • $begingroup$
      Thanks for responding.Got it.
      $endgroup$
      – NoChance
      31 mins ago




















    • $begingroup$
      Thanks,I got it.
      $endgroup$
      – Hitendra Kumar
      48 mins ago










    • $begingroup$
      You're welcome!
      $endgroup$
      – Stefan Lafon
      47 mins ago










    • $begingroup$
      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
      47 mins ago






    • 1




      $begingroup$
      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
      41 mins ago










    • $begingroup$
      Thanks for responding.Got it.
      $endgroup$
      – NoChance
      31 mins ago


















    $begingroup$
    Thanks,I got it.
    $endgroup$
    – Hitendra Kumar
    48 mins ago




    $begingroup$
    Thanks,I got it.
    $endgroup$
    – Hitendra Kumar
    48 mins ago












    $begingroup$
    You're welcome!
    $endgroup$
    – Stefan Lafon
    47 mins ago




    $begingroup$
    You're welcome!
    $endgroup$
    – Stefan Lafon
    47 mins ago












    $begingroup$
    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
    47 mins ago




    $begingroup$
    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
    47 mins ago




    1




    1




    $begingroup$
    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
    41 mins ago




    $begingroup$
    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
    41 mins ago












    $begingroup$
    Thanks for responding.Got it.
    $endgroup$
    – NoChance
    31 mins ago






    $begingroup$
    Thanks for responding.Got it.
    $endgroup$
    – NoChance
    31 mins ago













    3












    $begingroup$

    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
      1 hour ago
















    3












    $begingroup$

    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
      1 hour ago














    3












    3








    3





    $begingroup$

    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 1 hour ago









    David G. StorkDavid G. Stork

    11.1k41432




    11.1k41432












    • $begingroup$
      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
      1 hour ago


















    • $begingroup$
      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
      1 hour ago
















    $begingroup$
    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
    1 hour ago




    $begingroup$
    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
    1 hour ago











    0












    $begingroup$

    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 $n$. 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









    $endgroup$


















      0












      $begingroup$

      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 $n$. 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









      $endgroup$
















        0












        0








        0





        $begingroup$

        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 $n$. 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 $n$. 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










        answered 12 mins ago









        Mike EarnestMike Earnest

        25.5k22151




        25.5k22151






























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            Алькесар

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            Муниципалитет Алькесар Alquézar Герб 42°10′16″ с. ш. 0°01′27″ в. д. H G Я O L Страна Испания   Испания Автономное сообщество Арагон Провинция Уэска Комарка Сомонтано-де-Барбастро Алькальд Хосе Марияно Альтемир Ласкорс (ИСРП) История и география Площадь 32,36 км² Высота НУМ 660 м Часовой пояс UTC+1, летом UTC+2 Население Население 313 человек ( 2010 ) Плотность 9,92 чел./км² Этнохороним alquezarano/-a Цифровые идентификаторы Телефонный код +34 974 Почтовые индексы 22112 Автомобильный код HU alquezar.org   (исп.) Показать/скрыть карты Алькесар Алькесар Алькесар  Аудио, фото и видео на Викискладе Алькесар (исп.  Alquézar ) — муниципалитет в Испании, входит в провинцию Уэска, в составе автономного сообщества Арагон. Муниципалитет находится в составе района (комарки) Сомонтано-де-Барбастро. Занимает площадь 32,36 км². Население — 321 челове
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            Josef Freinademetz

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            San Ujöp (Giuseppe/Josef) Freinademetz Josef Freinademetz in abiti cinesi   Religioso e missionario   Nascita 15 aprile 1852 Morte 28 gennaio 1908 Venerato da Chiesa cattolica Beatificazione 1975 Canonizzazione 5 ottobre 2003 Ricorrenza 28 gennaio Manuale Ujöp Freinademetz noto anche come Giuseppe o Josef Freinademetz (Oies, 15 aprile 1852 – Taikia, 28 gennaio 1908) è stato un presbitero austriaco, membro della Società del Verbo Divino, fu missionario in Cina; è venerato come santo dalla Chiesa cattolica. .mw-parser-output .citazione-table{margin-bottom:.5em;font-size:95%}.mw-parser-output .citazione-table td{padding:0 1.2em 0 2.4em}.mw-parser-output .citazione-lang{vertical-align:top}.mw-parser-output .citazione-lang td{width:50%}.mw-parser-output .citazione-lang td:first-child{padding:0 0 0 2.4em}.mw-parser-output .citazione-lang td:nth-child(2){padding:0 1.2em} «Io amo la Cina e i cinesi; voglio morire in mezzo a loro, e tra loro
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