Sfrgttk

How do you make your own symbol when Detexify fails?

Showing a sum is positive

C++ check if statement can be evaluated constexpr

Mimic lecturing on blackboard, facing audience

What does Apple's new App Store requirement mean

Is this toilet slogan correct usage of the English language?

Which Article Helped Get Rid of Technobabble in RPGs?

How can I write humor as character trait?

Were Persian-Median kings illiterate?

Shouldn’t conservatives embrace universal basic income?

Why do ¬, ∀ and ∃ have the same precedence?

C++ copy constructor called at return

How does electrical safety system work on ISS?

Why is so much work done on numerical verification of the Riemann Hypothesis?

Why do Radio Buttons not fill the entire outer circle?

How many arrows is an archer expected to fire by the end of the Tyranny of Dragons pair of adventures?

What features enable the Su-25 Frogfoot to operate with such a wide variety of fuels?

What does "Scientists rise up against statistical significance" mean? (Comment in Nature)

It grows, but water kills it

What (the heck) is a Super Worm Equinox Moon?

Why is it that I can sometimes guess the next note?

How much theory knowledge is actually used while playing?

Do we have to expect a queue for the shuttle from Watford Junction to Harry Potter Studio?

What is the highest possible scrabble score for placing a single tile

Showing a sum is positive

Finding Binomial expansion of a radicalSimplify the Expression $sum _{ k=0 }^{ n }{ binom{n}{k}}i^{k}3^{k-n} $Alternative combinatorial proof for $sumlimits_{r=0}^nbinom{n}{r}binom{m+r}{n}=sumlimits_{r=0}^nbinom{n}{r}binom{m}{r}2^r$Proof by induction, binomial coefficientApproximating a binomial sum over a simplexHow to expand $sqrt{x^6+1}$ using Maclaurin's seriesSum of ${m choose j}$ multiplied by $2^{2^j}$How to show that $sumlimits_{k=0}^n (-1)^ktfrac{{ {n}choose{k}}}{{ {x+k}choose{k}}} = frac{x}{x+n}$Finite sum with inverse binomialShowing an alternating sum is positive

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

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$$sum_{k=0}^{n} {n choose k}frac{{(-1)}^k}{n+k+1}$$ 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

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$$sum_{k=0}^{n} {n choose k}frac{{(-1)}^k}{n+k+1}$$ 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

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:
$$begin{split}
sum_{k=0}^{n} {nchoose k}frac{{(-1)}^k}{n+k+1} &=sum_{k=0}^{n} {nchoose k}(-1)^kint_0^1 x^{n+k}dx\
&=int_0^1x^nsum_{k=0}^{n} {nchoose k}(-x)^kdx\
&=int_0^1x^n(1-x)^ndx
end{split}$$
The latter is clearly a positive number.

share|cite|improve this answer

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
  39 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
  38 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
  33 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for responding.Got it.
  $endgroup$
  – NoChance
  23 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
$$
boxed{sum_{k=0}^{n} {n choose k}frac{{(-1)}^k}{n+k+1}=left((2n+1)binom{2n}nright)^{-1}}
$$
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

answered 4 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

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

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
  39 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
  38 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
  33 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

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
  39 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
  38 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
  33 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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

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
$$
boxed{sum_{k=0}^{n} {n choose k}frac{{(-1)}^k}{n+k+1}=left((2n+1)binom{2n}nright)^{-1}}
$$
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

answered 4 mins ago

Mike EarnestMike Earnest

25.5k22151

$endgroup$

add a comment | 

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

answered 4 mins ago

Mike EarnestMike Earnest

25.5k22151

$endgroup$

add a comment | 

$begingroup$

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

answered 4 mins ago

Mike EarnestMike Earnest

25.5k22151

$endgroup$

share|cite

answered 4 mins ago

Mike EarnestMike Earnest

25.5k22151

answered 4 mins ago

Mike EarnestMike Earnest

25.5k22151

answered 4 mins ago

Mike EarnestMike Earnest

25.5k22151