Yhyjyk

"It doesn't matter" or "it won't matter"?

Which Article Helped Get Rid of Technobabble in RPGs?

How does electrical safety system work on ISS?

The Digit Triangles

Short story about a deaf man, who cuts people tongues

Is it allowed to activate the ability of multiple planeswalkers in a single turn?

Strong empirical falsification of quantum mechanics based on vacuum energy density?

It grows, but water kills it

Does "he squandered his car on drink" sound natural?

How to convince somebody that he is fit for something else, but not this job?

How to explain what's wrong with this application of the chain rule?

15% tax on $7.5k earnings. Is that right?

Is this toilet slogan correct usage of the English language?

The IT department bottlenecks progress, how should I handle this?

Quoting Keynes in a lecture

Has any country ever had 2 former presidents in jail simultaneously?

Why Shazam when there is already Superman?

I found an audio circuit and I built it just fine, but I find it a bit too quiet. How do I amplify the output so that it is a bit louder?

Is this part of the description of the Archfey warlock's Misty Escape feature redundant?

Stack Interview Code methods made from class Node and Smart Pointers

Plot of a tornado shape like surface

What is the English pronunciation of "pain au chocolat"?

What is going on with gets(stdin) on the site coderbyte?

How to make money from a browser who sees 5 seconds into the future of any web page?

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

3

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

656

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

3

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

656

$endgroup$

• 
  
  
  

  
  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

656

$endgroup$

share|cite|improve this question

edited 1 hour ago

Hitendra Kumar

asked 1 hour ago

Hitendra KumarHitendra Kumar

656

share|cite|improve this question

edited 1 hour ago

Hitendra Kumar

asked 1 hour ago

Hitendra KumarHitendra Kumar

656

asked 1 hour ago

Hitendra KumarHitendra Kumar

656

asked 1 hour ago

Hitendra KumarHitendra Kumar

656

3 Answers
3

active

oldest

votes

6

$begingroup$

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

3,02019

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks,I got it.
  $endgroup$
  – Hitendra Kumar
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome!
  $endgroup$
  – Stefan Lafon
  1 hour 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
  1 hour 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
  56 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for responding.Got it.
  $endgroup$
  – NoChance
  45 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 | 

1

$begingroup$

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|improve this answer

answered 26 mins ago

Mike EarnestMike Earnest

25.5k22151

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3157740%2fshowing-a-sum-is-positive%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

6

$begingroup$

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

3,02019

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks,I got it.
  $endgroup$
  – Hitendra Kumar
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome!
  $endgroup$
  – Stefan Lafon
  1 hour 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
  1 hour 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
  56 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

6

$begingroup$

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

3,02019

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks,I got it.
  $endgroup$
  – Hitendra Kumar
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome!
  $endgroup$
  – Stefan Lafon
  1 hour 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
  1 hour 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
  56 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for responding.Got it.
  $endgroup$
  – NoChance
  45 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

3,02019

$endgroup$

share|cite|improve this answer

answered 1 hour ago

Stefan LafonStefan Lafon

3,02019

answered 1 hour ago

Stefan LafonStefan Lafon

3,02019

answered 1 hour ago

Stefan LafonStefan Lafon

3,02019

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

1

$begingroup$

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|improve this answer

answered 26 mins ago

Mike EarnestMike Earnest

25.5k22151

$endgroup$

add a comment | 

1

$begingroup$

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|improve this answer

answered 26 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|improve this answer

answered 26 mins ago

Mike EarnestMike Earnest

25.5k22151

$endgroup$

share|cite|improve this answer

answered 26 mins ago

Mike EarnestMike Earnest

25.5k22151

answered 26 mins ago

Mike EarnestMike Earnest

25.5k22151

answered 26 mins ago

Mike EarnestMike Earnest

25.5k22151