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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

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

edited 1 hour ago

Hitendra Kumar

asked 1 hour ago

Hitendra KumarHitendra Kumar

606

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

edited 1 hour ago

Hitendra Kumar

asked 1 hour ago

Hitendra KumarHitendra Kumar

606

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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

edited 1 hour ago

Hitendra Kumar

asked 1 hour ago

Hitendra KumarHitendra Kumar

606

$endgroup$

share|cite|improve this question

edited 1 hour ago

Hitendra Kumar

asked 1 hour ago

Hitendra KumarHitendra Kumar

606

share|cite|improve this question

edited 1 hour ago

Hitendra Kumar

asked 1 hour ago

Hitendra KumarHitendra Kumar

606

asked 1 hour ago

Hitendra KumarHitendra Kumar

606

asked 1 hour ago

Hitendra KumarHitendra Kumar

606

3 Answers
3

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

answered 1 hour ago

Stefan LafonStefan Lafon

3,00019

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome!
  $endgroup$
  – Stefan Lafon
  40 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
  39 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
  34 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

answered 1 hour ago

David G. StorkDavid G. Stork

11.1k41432

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

0

$begingroup$

We can specifically prove that
$$
boxedsum_k=0^n n choose kfrac(-1)^kn+k+1=left((2n+1)binom2nnright)^-1
$$
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 $binomn1frac1n+2$. 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

answered 5 mins ago

Mike EarnestMike Earnest

25.5k22151

$endgroup$

add a comment | 

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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

answered 1 hour ago

Stefan LafonStefan Lafon

3,00019

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome!
  $endgroup$
  – Stefan Lafon
  40 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
  39 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
  34 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

answered 1 hour ago

Stefan LafonStefan Lafon

3,00019

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome!
  $endgroup$
  – Stefan Lafon
  40 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
  39 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
  34 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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

answered 1 hour ago

Stefan LafonStefan Lafon

3,00019

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Stefan LafonStefan Lafon

3,00019

answered 1 hour ago

Stefan LafonStefan Lafon

3,00019

answered 1 hour ago

Stefan LafonStefan Lafon

3,00019

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

answered 1 hour ago

David G. StorkDavid G. Stork

11.1k41432

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

answered 1 hour ago

David G. StorkDavid G. Stork

11.1k41432

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

answered 1 hour ago

David G. StorkDavid G. Stork

11.1k41432

$endgroup$

share|cite|improve this answer

answered 1 hour ago

David G. StorkDavid G. Stork

11.1k41432

answered 1 hour ago

David G. StorkDavid G. Stork

11.1k41432

answered 1 hour ago

David G. StorkDavid G. Stork

11.1k41432

0

$begingroup$

We can specifically prove that
$$
boxedsum_k=0^n n choose kfrac(-1)^kn+k+1=left((2n+1)binom2nnright)^-1
$$
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 $binomn1frac1n+2$. 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

answered 5 mins ago

Mike EarnestMike Earnest

25.5k22151

$endgroup$

add a comment | 

0

$begingroup$

We can specifically prove that
$$
boxedsum_k=0^n n choose kfrac(-1)^kn+k+1=left((2n+1)binom2nnright)^-1
$$
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 $binomn1frac1n+2$. 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

answered 5 mins ago

Mike EarnestMike Earnest

25.5k22151

$endgroup$

add a comment | 

$begingroup$

We can specifically prove that
$$
boxedsum_k=0^n n choose kfrac(-1)^kn+k+1=left((2n+1)binom2nnright)^-1
$$
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 $binomn1frac1n+2$. 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

answered 5 mins ago

Mike EarnestMike Earnest

25.5k22151

$endgroup$

share|cite

answered 5 mins ago

Mike EarnestMike Earnest

25.5k22151

answered 5 mins ago

Mike EarnestMike Earnest

25.5k22151

answered 5 mins ago

Mike EarnestMike Earnest

25.5k22151