Showing a sum is positive












3












$begingroup$



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


















3












$begingroup$



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
















3












3








3


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

656




656








  • 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


















6












$begingroup$

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
    54 mins ago










  • $begingroup$
    You're welcome!
    $endgroup$
    – Stefan Lafon
    53 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
    52 mins ago






  • 2




    $begingroup$
    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
    47 mins ago










  • $begingroup$
    Thanks for responding.Got it.
    $endgroup$
    – NoChance
    37 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



















1












$begingroup$

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









    6












    $begingroup$

    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
      54 mins ago










    • $begingroup$
      You're welcome!
      $endgroup$
      – Stefan Lafon
      53 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
      52 mins ago






    • 2




      $begingroup$
      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
      47 mins ago










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


















    6












    $begingroup$

    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
      54 mins ago










    • $begingroup$
      You're welcome!
      $endgroup$
      – Stefan Lafon
      53 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
      52 mins ago






    • 2




      $begingroup$
      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
      47 mins ago










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
















    6












    6








    6





    $begingroup$

    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

    3,02019




    3,02019












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










    • $begingroup$
      You're welcome!
      $endgroup$
      – Stefan Lafon
      53 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
      52 mins ago






    • 2




      $begingroup$
      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
      47 mins ago










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




















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










    • $begingroup$
      You're welcome!
      $endgroup$
      – Stefan Lafon
      53 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
      52 mins ago






    • 2




      $begingroup$
      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
      47 mins ago










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


















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




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












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




    $begingroup$
    You're welcome!
    $endgroup$
    – Stefan Lafon
    53 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
    52 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
    52 mins ago




    2




    2




    $begingroup$
    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
    47 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
    47 mins ago












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






    $begingroup$
    Thanks for responding.Got it.
    $endgroup$
    – NoChance
    37 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











    1












    $begingroup$

    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









    $endgroup$


















      1












      $begingroup$

      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









      $endgroup$
















        1












        1








        1





        $begingroup$

        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 18 mins ago









        Mike EarnestMike Earnest

        25.5k22151




        25.5k22151






























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            الفوسفات في المغرب

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            قطار ينقل فوسفات من منطقة خريبكة متجه نحو الدار البيضاء. ديار عمال بمصانع الفوسفات بمدينة آسفي. المحتوى هنا ينقصه الاستشهاد بمصادر. يرجى إيراد مصادر موثوق بها. أي معلومات غير موثقة يمكن التشكيك بها وإزالتها. (أغسطس 2015) الفوسفات في المغرب هو ثالث مصدر عالمي للفوسفات، ويتوفر المغرب على %75 من الاحتياطي العالمي  [بحاجة لمصدر] ويتوزع بنسب متفاوتة بين عدة مناطق أساسية وهي: وادي زم بن جرير بوكراع خريبكة اليوسفية أولاد عبدون كنتور واد الذهب مستقلة. هذه الأخيرة لم تفتح بعد للاستغلال. الصادرات الحامض الفوسفوري P205 تمتص الهند %50 من ا لصادرات الإجمالية آسيا (دون روسيا)%59 أوروبا %30 أمريكا اللاتينية %11 أسمدة صلبة أوروبا 38 % أمريكا اللاتينية %30 آسيا (دون روسيا) 22 % أستراليا 9 % بوابة المغرب هذه بذرة مقالة عن علم الاقتصاد أو موضوع متعلق به بحاجة للتوسيع. شارك في تحريرها. This page is only for reference, If you need detailed information, please check here
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            Four equal circles intersect: What is the area of the small shaded portion and its height

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            3 $begingroup$ In terms of $R$ which is the radius of all four circles, what is the area of the intersection region of these four equal circles and the height of the marked arrow in the figure? The marked arrow is along the line CD, also the midpoint of all the circles are points A, B, C and D. Looking for a very short intuitive solution. I have checked similar questions on this site for example this and this. geometry circle share | cite | improve this question edited 2 hours ago Abdulhameed asked 2 hours ago ...
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            جامعة ليفربول

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            جامعة ليفربول University of Liverpool الأسماء السابقة كلية ليفربول الجامعية (1882-1884) جامعة فيكتوريا في ليفربول (1884-1903) معلومات التأسيس 1882 (منذ 137 سنة) الإنتماءات رابطة الجامعات الأوروبية مجموعة راسل جامعات القرميد الأحمر المنحة المالية 143.5 مليون £ النوع جامعة عامة الكليات 6 الموقع الجغرافي إحداثيات 53°24′22″N 2°58′01″W  /  53.406°N 2.967°W  / 53.406; -2.967   المدينة ليفربول المكان   إنجلترا البلد   المملكة المتحدة رقم الهاتف 7942000 151 (44+) إحصاءات الطلاب 17,405 (سنة 2008) عضوية مجموعة رسل [1] أورسيد [2]   طلاب الدراسات العليا 3,860 الموقع www.liverpool.ac.uk تعديل   جامعة ليفربول مبنى فكتوريا جامعة ليفربول (بالإنجليزية: The University of Liverpool )) إحدى الجامعات البريطانية العريقة، تقع في مدينة ليفربول على الساحل الشمالي الغربي من أنجلترا . وهي عضو في مجموعة راسل للجامعات وتعتبر من ضمن جامعات الطوب الأحمر الستة. محتويات ...
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