Showing a sum is positiveFinding Binomial expansion of a radicalSimplify the Expression $sum _ k=0 ^ n binomnki^k3^k-n $Alternative combinatorial proof for $sumlimits_r=0^nbinomnrbinomm+rn=sumlimits_r=0^nbinomnrbinommr2^r$Proof by induction, binomial coefficientApproximating a binomial sum over a simplexHow to expand $sqrtx^6+1$ using Maclaurin's seriesSum of $m choose j$ multiplied by $2^2^j$How to show that $sumlimits_k=0^n (-1)^ktfrac nchoosek x+kchoosek = fracxx+n$Finite sum with inverse binomialShowing an alternating sum is positive

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Showing a sum is positive


Finding Binomial expansion of a radicalSimplify the Expression $sum _ k=0 ^ n binomnki^k3^k-n $Alternative combinatorial proof for $sumlimits_r=0^nbinomnrbinomm+rn=sumlimits_r=0^nbinomnrbinommr2^r$Proof by induction, binomial coefficientApproximating a binomial sum over a simplexHow to expand $sqrtx^6+1$ using Maclaurin's seriesSum of $m choose j$ multiplied by $2^2^j$How to show that $sumlimits_k=0^n (-1)^ktfrac nchoosek x+kchoosek = fracxx+n$Finite sum with inverse binomialShowing an alternating sum is positive













2












$begingroup$



Show that the sum$$sum_k=0^n n choose kfrac(-1)^kn+k+1$$ 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$$sum_k=0^n n choose kfrac(-1)^kn+k+1$$ 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$$sum_k=0^n n choose kfrac(-1)^kn+k+1$$ 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$$sum_k=0^n n choose kfrac(-1)^kn+k+1$$ 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











2 Answers
2






active

oldest

votes


















4












$begingroup$

Direct proof:
$$beginsplit
sum_k=0^n nchoose kfrac(-1)^kn+k+1 &=sum_k=0^n nchoose k(-1)^kint_0^1 x^n+kdx\
&=int_0^1x^nsum_k=0^n nchoose k(-x)^kdx\
&=int_0^1x^n(1-x)^ndx
endsplit$$
The latter is clearly a positive number.






share|cite|improve this answer









$endgroup$












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










  • $begingroup$
    You're welcome!
    $endgroup$
    – Stefan Lafon
    34 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
    34 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
    28 mins ago










  • $begingroup$
    Thanks for responding.Got it.
    $endgroup$
    – NoChance
    18 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










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Direct proof:
$$beginsplit
sum_k=0^n nchoose kfrac(-1)^kn+k+1 &=sum_k=0^n nchoose k(-1)^kint_0^1 x^n+kdx\
&=int_0^1x^nsum_k=0^n nchoose k(-x)^kdx\
&=int_0^1x^n(1-x)^ndx
endsplit$$
The latter is clearly a positive number.






share|cite|improve this answer









$endgroup$












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










  • $begingroup$
    You're welcome!
    $endgroup$
    – Stefan Lafon
    34 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
    34 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
    28 mins ago










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
















4












$begingroup$

Direct proof:
$$beginsplit
sum_k=0^n nchoose kfrac(-1)^kn+k+1 &=sum_k=0^n nchoose k(-1)^kint_0^1 x^n+kdx\
&=int_0^1x^nsum_k=0^n nchoose k(-x)^kdx\
&=int_0^1x^n(1-x)^ndx
endsplit$$
The latter is clearly a positive number.






share|cite|improve this answer









$endgroup$












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










  • $begingroup$
    You're welcome!
    $endgroup$
    – Stefan Lafon
    34 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
    34 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
    28 mins ago










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














4












4








4





$begingroup$

Direct proof:
$$beginsplit
sum_k=0^n nchoose kfrac(-1)^kn+k+1 &=sum_k=0^n nchoose k(-1)^kint_0^1 x^n+kdx\
&=int_0^1x^nsum_k=0^n nchoose k(-x)^kdx\
&=int_0^1x^n(1-x)^ndx
endsplit$$
The latter is clearly a positive number.






share|cite|improve this answer









$endgroup$



Direct proof:
$$beginsplit
sum_k=0^n nchoose kfrac(-1)^kn+k+1 &=sum_k=0^n nchoose k(-1)^kint_0^1 x^n+kdx\
&=int_0^1x^nsum_k=0^n nchoose k(-x)^kdx\
&=int_0^1x^n(1-x)^ndx
endsplit$$
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
    35 mins ago










  • $begingroup$
    You're welcome!
    $endgroup$
    – Stefan Lafon
    34 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
    34 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
    28 mins ago










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

















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










  • $begingroup$
    You're welcome!
    $endgroup$
    – Stefan Lafon
    34 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
    34 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
    28 mins ago










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
















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




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












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




$begingroup$
You're welcome!
$endgroup$
– Stefan Lafon
34 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
34 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
34 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
28 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
28 mins ago












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





$begingroup$
Thanks for responding.Got it.
$endgroup$
– NoChance
18 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

















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