Mixing
Conjecture: $sumlimits_{ngeq0}left(frac12right)^nprodlimits_{k=1}^{n}frac{2n-2k+1}{2n-2k+2}=sqrt2$...
$begingroup$
This question already has an answer here:
Summing the power series $sumlimits_{n=0}^infty (-1)^n frac{x^{2n+1}}{2n+1}prodlimits_{k=1}^nfrac{2k-1}{2k} $
4 answers
I am trying to solve a problem I made form myself: proving that
$$sum_{ngeq0}left(frac12right)^nprod_{k=1}^{n}frac{2n-2k+1}{2n-2k+2}=sqrt2$$
The highly accurate powers of Desmos seem to confirm my hunch. But how do I prove it?
I was just messing around with products and generating functions, and then I noticed that the numerical value of the sum in question was suspiciously similar to $sqrt2$, so I conjectured the result. Unfortunately I found this completely by accident and have absolutely no idea of how to prove it.
Feeble attempt:
Define $$S(x)=sum_{ngeq0}x^nprod_{k=1}^{n}frac{2n-2k+1}{2n-2k+2}$$
Which Wolfram says is equal to $$S(x)=sum_{ngeq0}x^nfrac{(1/2-n)_n}{(-n)_n}$$
With $displaystyle (x)_n=frac{Gamma(x+n)}{Gamma(x)}$. But that doesn't really make sense because $Gamma(0)$ is undefined. So all in all I'm just confused.
Could I have some help?
Edit:
According to the comments, it suffices to prove that
$$sum_{ngeq1}left(frac12right)^nprod_{k=1}^{n}frac{2n-2k+1}{2n-2k+2}=sqrt2-1$$
real-analysis sequences-and-series combinatorics closed-form
edited Jan 21 at 6:03
Martin Sleziak
44.8k10119273
asked Jan 20 at 22:25
clathratusclathratus
4,839338
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Jan 21 at 7:58
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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show 1 more comment
$begingroup$
This question already has an answer here:
Summing the power series $sumlimits_{n=0}^infty (-1)^n frac{x^{2n+1}}{2n+1}prodlimits_{k=1}^nfrac{2k-1}{2k} $
4 answers
I am trying to solve a problem I made form myself: proving that
$$sum_{ngeq0}left(frac12right)^nprod_{k=1}^{n}frac{2n-2k+1}{2n-2k+2}=sqrt2$$
The highly accurate powers of Desmos seem to confirm my hunch. But how do I prove it?
I was just messing around with products and generating functions, and then I noticed that the numerical value of the sum in question was suspiciously similar to $sqrt2$, so I conjectured the result. Unfortunately I found this completely by accident and have absolutely no idea of how to prove it.
Feeble attempt:
Define $$S(x)=sum_{ngeq0}x^nprod_{k=1}^{n}frac{2n-2k+1}{2n-2k+2}$$
Which Wolfram says is equal to $$S(x)=sum_{ngeq0}x^nfrac{(1/2-n)_n}{(-n)_n}$$
With $displaystyle (x)_n=frac{Gamma(x+n)}{Gamma(x)}$. But that doesn't really make sense because $Gamma(0)$ is undefined. So all in all I'm just confused.
Could I have some help?
Edit:
According to the comments, it suffices to prove that
$$sum_{ngeq1}left(frac12right)^nprod_{k=1}^{n}frac{2n-2k+1}{2n-2k+2}=sqrt2-1$$
real-analysis sequences-and-series combinatorics closed-form
edited Jan 21 at 6:03
Martin Sleziak
44.8k10119273
asked Jan 20 at 22:25
clathratusclathratus
4,839338
$endgroup$
marked as duplicate by Did real-analysis
Users with the real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.
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Jan 21 at 7:58
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Satisfy yourself that $(0)_0=1$.
$endgroup$
– J.G.
Jan 20 at 22:30
$begingroup$
@J.G. I am now satisfied
$endgroup$
– clathratus
Jan 20 at 22:30
$begingroup$
In case of $n=0$, product in your sum is empty product, so you may regard it as 1. (Or just do the sum for $ngeq1$)
$endgroup$
– Seewoo Lee
Jan 20 at 22:31
$begingroup$
I don't think you need to include the $n=0$ case because the product is not defined for it.
$endgroup$
– Antinous
Jan 20 at 22:31
$begingroup$
@Antinous For $n=0$ the product is empty, and thus conventionally defined to equal $1$
$endgroup$
– Learner
Jan 20 at 22:59
|
show 1 more comment
$begingroup$
This question already has an answer here:
Summing the power series $sumlimits_{n=0}^infty (-1)^n frac{x^{2n+1}}{2n+1}prodlimits_{k=1}^nfrac{2k-1}{2k} $
4 answers
I am trying to solve a problem I made form myself: proving that
$$sum_{ngeq0}left(frac12right)^nprod_{k=1}^{n}frac{2n-2k+1}{2n-2k+2}=sqrt2$$
The highly accurate powers of Desmos seem to confirm my hunch. But how do I prove it?
I was just messing around with products and generating functions, and then I noticed that the numerical value of the sum in question was suspiciously similar to $sqrt2$, so I conjectured the result. Unfortunately I found this completely by accident and have absolutely no idea of how to prove it.
Feeble attempt:
Define $$S(x)=sum_{ngeq0}x^nprod_{k=1}^{n}frac{2n-2k+1}{2n-2k+2}$$
Which Wolfram says is equal to $$S(x)=sum_{ngeq0}x^nfrac{(1/2-n)_n}{(-n)_n}$$
With $displaystyle (x)_n=frac{Gamma(x+n)}{Gamma(x)}$. But that doesn't really make sense because $Gamma(0)$ is undefined. So all in all I'm just confused.
Could I have some help?
Edit:
According to the comments, it suffices to prove that
$$sum_{ngeq1}left(frac12right)^nprod_{k=1}^{n}frac{2n-2k+1}{2n-2k+2}=sqrt2-1$$
real-analysis sequences-and-series combinatorics closed-form
edited Jan 21 at 6:03
Martin Sleziak
44.8k10119273
asked Jan 20 at 22:25
clathratusclathratus
4,839338
$endgroup$
This question already has an answer here:
Summing the power series $sumlimits_{n=0}^infty (-1)^n frac{x^{2n+1}}{2n+1}prodlimits_{k=1}^nfrac{2k-1}{2k} $
4 answers
I am trying to solve a problem I made form myself: proving that
$$sum_{ngeq0}left(frac12right)^nprod_{k=1}^{n}frac{2n-2k+1}{2n-2k+2}=sqrt2$$
The highly accurate powers of Desmos seem to confirm my hunch. But how do I prove it?
I was just messing around with products and generating functions, and then I noticed that the numerical value of the sum in question was suspiciously similar to $sqrt2$, so I conjectured the result. Unfortunately I found this completely by accident and have absolutely no idea of how to prove it.
Feeble attempt:
Define $$S(x)=sum_{ngeq0}x^nprod_{k=1}^{n}frac{2n-2k+1}{2n-2k+2}$$
Which Wolfram says is equal to $$S(x)=sum_{ngeq0}x^nfrac{(1/2-n)_n}{(-n)_n}$$
With $displaystyle (x)_n=frac{Gamma(x+n)}{Gamma(x)}$. But that doesn't really make sense because $Gamma(0)$ is undefined. So all in all I'm just confused.
Could I have some help?
Edit:
According to the comments, it suffices to prove that
$$sum_{ngeq1}left(frac12right)^nprod_{k=1}^{n}frac{2n-2k+1}{2n-2k+2}=sqrt2-1$$
This question already has an answer here:
Summing the power series $sumlimits_{n=0}^infty (-1)^n frac{x^{2n+1}}{2n+1}prodlimits_{k=1}^nfrac{2k-1}{2k} $
4 answers
real-analysis sequences-and-series combinatorics closed-form
real-analysis sequences-and-series combinatorics closed-form
edited Jan 21 at 6:03
Martin Sleziak
44.8k10119273
asked Jan 20 at 22:25
clathratusclathratus
4,839338
edited Jan 21 at 6:03
Martin Sleziak
44.8k10119273
asked Jan 20 at 22:25
clathratusclathratus
4,839338
edited Jan 21 at 6:03
Martin Sleziak
44.8k10119273
edited Jan 21 at 6:03
Martin Sleziak
44.8k10119273
edited Jan 21 at 6:03
Martin Sleziak
44.8k10119273
44.8k10119273
asked Jan 20 at 22:25
clathratusclathratus
4,839338
asked Jan 20 at 22:25
clathratusclathratus
4,839338
asked Jan 20 at 22:25
clathratusclathratus
4,839338
4,839338
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Jan 21 at 7:58
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Did real-analysis
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Jan 21 at 7:58
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Satisfy yourself that $(0)_0=1$.
$endgroup$
– J.G.
Jan 20 at 22:30
$begingroup$
@J.G. I am now satisfied
$endgroup$
– clathratus
Jan 20 at 22:30
$begingroup$
In case of $n=0$, product in your sum is empty product, so you may regard it as 1. (Or just do the sum for $ngeq1$)
$endgroup$
– Seewoo Lee
Jan 20 at 22:31
$begingroup$
I don't think you need to include the $n=0$ case because the product is not defined for it.
$endgroup$
– Antinous
Jan 20 at 22:31
$begingroup$
@Antinous For $n=0$ the product is empty, and thus conventionally defined to equal $1$
$endgroup$
– Learner
Jan 20 at 22:59
|
show 1 more comment
$begingroup$
Satisfy yourself that $(0)_0=1$.
$endgroup$
– J.G.
Jan 20 at 22:30
$begingroup$
@J.G. I am now satisfied
$endgroup$
– clathratus
Jan 20 at 22:30
$begingroup$
In case of $n=0$, product in your sum is empty product, so you may regard it as 1. (Or just do the sum for $ngeq1$)
$endgroup$
– Seewoo Lee
Jan 20 at 22:31
$begingroup$
I don't think you need to include the $n=0$ case because the product is not defined for it.
$endgroup$
– Antinous
Jan 20 at 22:31
$begingroup$
@Antinous For $n=0$ the product is empty, and thus conventionally defined to equal $1$
$endgroup$
– Learner
Jan 20 at 22:59
$begingroup$
Satisfy yourself that $(0)_0=1$.
$endgroup$
– J.G.
Jan 20 at 22:30
$begingroup$
Satisfy yourself that $(0)_0=1$.
$endgroup$
– J.G.
Jan 20 at 22:30
$begingroup$
@J.G. I am now satisfied
$endgroup$
– clathratus
Jan 20 at 22:30
$begingroup$
@J.G. I am now satisfied
$endgroup$
– clathratus
Jan 20 at 22:30
$begingroup$
In case of $n=0$, product in your sum is empty product, so you may regard it as 1. (Or just do the sum for $ngeq1$)
$endgroup$
– Seewoo Lee
Jan 20 at 22:31
$begingroup$
In case of $n=0$, product in your sum is empty product, so you may regard it as 1. (Or just do the sum for $ngeq1$)
$endgroup$
– Seewoo Lee
Jan 20 at 22:31
$begingroup$
I don't think you need to include the $n=0$ case because the product is not defined for it.
$endgroup$
– Antinous
Jan 20 at 22:31
$begingroup$
I don't think you need to include the $n=0$ case because the product is not defined for it.
$endgroup$
– Antinous
Jan 20 at 22:31
$begingroup$
@Antinous For $n=0$ the product is empty, and thus conventionally defined to equal $1$
$endgroup$
– Learner
Jan 20 at 22:59
$begingroup$
@Antinous For $n=0$ the product is empty, and thus conventionally defined to equal $1$
$endgroup$
– Learner
Jan 20 at 22:59
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
The products can be rewritten as
begin{eqnarray*}
sum_{n=0}^{infty} frac{(2n-1)!!}{(2n)!!} frac{1}{2^n} = sum_{n=0}^{infty} binom{2n}{n} frac{1}{8^n}
end{eqnarray*}
Now use
begin{eqnarray*}
sum_{n=0}^{infty} binom{2n}{n} x^n = frac{1}{sqrt{1-4x}}.
end{eqnarray*}
Edit:
begin{eqnarray*}
prod_{k=1}^{n}frac{2n-2k+1}{2n-2k+2}&=&prod_{k=1}^{n}frac{2k-1}{2k} =frac{(2n-1)!!}{(2n)!!}\
&=&frac{(2n)!}{(2^n n!)^2}=binom{2n}{n} frac{1}{2^{2n}}.
end{eqnarray*}
answered Jan 20 at 22:35
Donald SplutterwitDonald Splutterwit
22.9k21446
$endgroup$
1
$begingroup$
Woah seriously?? (+1) That was fast! By the way, could you include the proof for your magical rewriting of products?
$endgroup$
– clathratus
Jan 20 at 22:36
$begingroup$
Wait actually no: I'm not gonna accept your answer until you show in detail how $$prod_{k=1}^{n}frac{2n-2k+1}{2n-2k+2}=frac{(2n-1)!!}{(2n)!!}=frac1{2^{2n}}{2nchoose n}$$
$endgroup$
– clathratus
Jan 20 at 22:47
$begingroup$
Sure ... give me a few mins ?
$endgroup$
– Donald Splutterwit
Jan 20 at 22:48
$begingroup$
Very nice. Thank you!
$endgroup$
– clathratus
Jan 20 at 23:42
1
$begingroup$
To add some kind of reference for the result on $binom{2n}n$, I'll mention links some other posts on this site: Generating functions and central binomial coefficient and Show $sumlimits_{n=0}^{infty}{2n choose n}x^n=(1-4x)^{-1/2}$. (And probably more posts about this can be found.)
$endgroup$
– Martin Sleziak
Jan 21 at 6:06
add a comment |
$begingroup$
Another way is
$$
eqalign{
& prodlimits_{k = 1}^n {{{2n - 2k + 1} over {2n - 2k + 2}}} = prodlimits_{k = 1}^n {{{left( {n - k} right) + 1/2} over {left( {n - k} right) + 1}}}
= prodlimits_{j = 0}^{n - 1} {{{j + 1/2} over {j + 1}}} = cr
& = {{left( {1/2} right)^{,overline {,n,} } } over {n!}}
= {{Gamma left( {n + 1/2} right)} over {Gamma left( {1/2} right)Gamma left( {n + 1} right)}} =
left( matrix{ n - 1/2 cr n cr} right)
= left( { - 1} right)^{,n} left( matrix{ - 1/2 cr n cr} right) cr}
$$
and then apply the binomial expansion.
answered Jan 20 at 23:47
G CabG Cab
19.8k31340
$endgroup$
$begingroup$
Neat! thank you!
$endgroup$
– clathratus
Jan 20 at 23:57
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The products can be rewritten as
begin{eqnarray*}
sum_{n=0}^{infty} frac{(2n-1)!!}{(2n)!!} frac{1}{2^n} = sum_{n=0}^{infty} binom{2n}{n} frac{1}{8^n}
end{eqnarray*}
Now use
begin{eqnarray*}
sum_{n=0}^{infty} binom{2n}{n} x^n = frac{1}{sqrt{1-4x}}.
end{eqnarray*}
Edit:
begin{eqnarray*}
prod_{k=1}^{n}frac{2n-2k+1}{2n-2k+2}&=&prod_{k=1}^{n}frac{2k-1}{2k} =frac{(2n-1)!!}{(2n)!!}\
&=&frac{(2n)!}{(2^n n!)^2}=binom{2n}{n} frac{1}{2^{2n}}.
end{eqnarray*}
answered Jan 20 at 22:35
Donald SplutterwitDonald Splutterwit
22.9k21446
$endgroup$
1
$begingroup$
Woah seriously?? (+1) That was fast! By the way, could you include the proof for your magical rewriting of products?
$endgroup$
– clathratus
Jan 20 at 22:36
$begingroup$
Wait actually no: I'm not gonna accept your answer until you show in detail how $$prod_{k=1}^{n}frac{2n-2k+1}{2n-2k+2}=frac{(2n-1)!!}{(2n)!!}=frac1{2^{2n}}{2nchoose n}$$
$endgroup$
– clathratus
Jan 20 at 22:47
$begingroup$
Sure ... give me a few mins ?
$endgroup$
– Donald Splutterwit
Jan 20 at 22:48
$begingroup$
Very nice. Thank you!
$endgroup$
– clathratus
Jan 20 at 23:42
1
$begingroup$
To add some kind of reference for the result on $binom{2n}n$, I'll mention links some other posts on this site: Generating functions and central binomial coefficient and Show $sumlimits_{n=0}^{infty}{2n choose n}x^n=(1-4x)^{-1/2}$. (And probably more posts about this can be found.)
$endgroup$
– Martin Sleziak
Jan 21 at 6:06
add a comment |
$begingroup$
The products can be rewritten as
begin{eqnarray*}
sum_{n=0}^{infty} frac{(2n-1)!!}{(2n)!!} frac{1}{2^n} = sum_{n=0}^{infty} binom{2n}{n} frac{1}{8^n}
end{eqnarray*}
Now use
begin{eqnarray*}
sum_{n=0}^{infty} binom{2n}{n} x^n = frac{1}{sqrt{1-4x}}.
end{eqnarray*}
Edit:
begin{eqnarray*}
prod_{k=1}^{n}frac{2n-2k+1}{2n-2k+2}&=&prod_{k=1}^{n}frac{2k-1}{2k} =frac{(2n-1)!!}{(2n)!!}\
&=&frac{(2n)!}{(2^n n!)^2}=binom{2n}{n} frac{1}{2^{2n}}.
end{eqnarray*}
answered Jan 20 at 22:35
Donald SplutterwitDonald Splutterwit
22.9k21446
$endgroup$
1
$begingroup$
Woah seriously?? (+1) That was fast! By the way, could you include the proof for your magical rewriting of products?
$endgroup$
– clathratus
Jan 20 at 22:36
$begingroup$
Wait actually no: I'm not gonna accept your answer until you show in detail how $$prod_{k=1}^{n}frac{2n-2k+1}{2n-2k+2}=frac{(2n-1)!!}{(2n)!!}=frac1{2^{2n}}{2nchoose n}$$
$endgroup$
– clathratus
Jan 20 at 22:47
$begingroup$
Sure ... give me a few mins ?
$endgroup$
– Donald Splutterwit
Jan 20 at 22:48
$begingroup$
Very nice. Thank you!
$endgroup$
– clathratus
Jan 20 at 23:42
1
$begingroup$
To add some kind of reference for the result on $binom{2n}n$, I'll mention links some other posts on this site: Generating functions and central binomial coefficient and Show $sumlimits_{n=0}^{infty}{2n choose n}x^n=(1-4x)^{-1/2}$. (And probably more posts about this can be found.)
$endgroup$
– Martin Sleziak
Jan 21 at 6:06
add a comment |
$begingroup$
The products can be rewritten as
begin{eqnarray*}
sum_{n=0}^{infty} frac{(2n-1)!!}{(2n)!!} frac{1}{2^n} = sum_{n=0}^{infty} binom{2n}{n} frac{1}{8^n}
end{eqnarray*}
Now use
begin{eqnarray*}
sum_{n=0}^{infty} binom{2n}{n} x^n = frac{1}{sqrt{1-4x}}.
end{eqnarray*}
Edit:
begin{eqnarray*}
prod_{k=1}^{n}frac{2n-2k+1}{2n-2k+2}&=&prod_{k=1}^{n}frac{2k-1}{2k} =frac{(2n-1)!!}{(2n)!!}\
&=&frac{(2n)!}{(2^n n!)^2}=binom{2n}{n} frac{1}{2^{2n}}.
end{eqnarray*}
answered Jan 20 at 22:35
Donald SplutterwitDonald Splutterwit
22.9k21446
$endgroup$
The products can be rewritten as
begin{eqnarray*}
sum_{n=0}^{infty} frac{(2n-1)!!}{(2n)!!} frac{1}{2^n} = sum_{n=0}^{infty} binom{2n}{n} frac{1}{8^n}
end{eqnarray*}
Now use
begin{eqnarray*}
sum_{n=0}^{infty} binom{2n}{n} x^n = frac{1}{sqrt{1-4x}}.
end{eqnarray*}
Edit:
begin{eqnarray*}
prod_{k=1}^{n}frac{2n-2k+1}{2n-2k+2}&=&prod_{k=1}^{n}frac{2k-1}{2k} =frac{(2n-1)!!}{(2n)!!}\
&=&frac{(2n)!}{(2^n n!)^2}=binom{2n}{n} frac{1}{2^{2n}}.
end{eqnarray*}
answered Jan 20 at 22:35
Donald SplutterwitDonald Splutterwit
22.9k21446
edited Jan 20 at 22:55
answered Jan 20 at 22:35
Donald SplutterwitDonald Splutterwit
22.9k21446
answered Jan 20 at 22:35
Donald SplutterwitDonald Splutterwit
22.9k21446
answered Jan 20 at 22:35
Donald SplutterwitDonald Splutterwit
22.9k21446
22.9k21446
1
$begingroup$
Woah seriously?? (+1) That was fast! By the way, could you include the proof for your magical rewriting of products?
$endgroup$
– clathratus
Jan 20 at 22:36
$begingroup$
Wait actually no: I'm not gonna accept your answer until you show in detail how $$prod_{k=1}^{n}frac{2n-2k+1}{2n-2k+2}=frac{(2n-1)!!}{(2n)!!}=frac1{2^{2n}}{2nchoose n}$$
$endgroup$
– clathratus
Jan 20 at 22:47
$begingroup$
Sure ... give me a few mins ?
$endgroup$
– Donald Splutterwit
Jan 20 at 22:48
$begingroup$
Very nice. Thank you!
$endgroup$
– clathratus
Jan 20 at 23:42
1
$begingroup$
To add some kind of reference for the result on $binom{2n}n$, I'll mention links some other posts on this site: Generating functions and central binomial coefficient and Show $sumlimits_{n=0}^{infty}{2n choose n}x^n=(1-4x)^{-1/2}$. (And probably more posts about this can be found.)
$endgroup$
– Martin Sleziak
Jan 21 at 6:06
add a comment |
1
$begingroup$
Woah seriously?? (+1) That was fast! By the way, could you include the proof for your magical rewriting of products?
$endgroup$
– clathratus
Jan 20 at 22:36
$begingroup$
Wait actually no: I'm not gonna accept your answer until you show in detail how $$prod_{k=1}^{n}frac{2n-2k+1}{2n-2k+2}=frac{(2n-1)!!}{(2n)!!}=frac1{2^{2n}}{2nchoose n}$$
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– clathratus
Jan 20 at 22:47
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Sure ... give me a few mins ?
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– Donald Splutterwit
Jan 20 at 22:48
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Very nice. Thank you!
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– clathratus
Jan 20 at 23:42
1
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To add some kind of reference for the result on $binom{2n}n$, I'll mention links some other posts on this site: Generating functions and central binomial coefficient and Show $sumlimits_{n=0}^{infty}{2n choose n}x^n=(1-4x)^{-1/2}$. (And probably more posts about this can be found.)
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– Martin Sleziak
Jan 21 at 6:06
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1
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Woah seriously?? (+1) That was fast! By the way, could you include the proof for your magical rewriting of products?
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– clathratus
Jan 20 at 22:36
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Woah seriously?? (+1) That was fast! By the way, could you include the proof for your magical rewriting of products?
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– clathratus
Jan 20 at 22:36
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Wait actually no: I'm not gonna accept your answer until you show in detail how $$prod_{k=1}^{n}frac{2n-2k+1}{2n-2k+2}=frac{(2n-1)!!}{(2n)!!}=frac1{2^{2n}}{2nchoose n}$$
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– clathratus
Jan 20 at 22:47
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Wait actually no: I'm not gonna accept your answer until you show in detail how $$prod_{k=1}^{n}frac{2n-2k+1}{2n-2k+2}=frac{(2n-1)!!}{(2n)!!}=frac1{2^{2n}}{2nchoose n}$$
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– clathratus
Jan 20 at 22:47
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Sure ... give me a few mins ?
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– Donald Splutterwit
Jan 20 at 22:48
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Sure ... give me a few mins ?
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– Donald Splutterwit
Jan 20 at 22:48
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Very nice. Thank you!
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– clathratus
Jan 20 at 23:42
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Very nice. Thank you!
$endgroup$
– clathratus
Jan 20 at 23:42
1
1
$begingroup$
To add some kind of reference for the result on $binom{2n}n$, I'll mention links some other posts on this site: Generating functions and central binomial coefficient and Show $sumlimits_{n=0}^{infty}{2n choose n}x^n=(1-4x)^{-1/2}$. (And probably more posts about this can be found.)
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– Martin Sleziak
Jan 21 at 6:06
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To add some kind of reference for the result on $binom{2n}n$, I'll mention links some other posts on this site: Generating functions and central binomial coefficient and Show $sumlimits_{n=0}^{infty}{2n choose n}x^n=(1-4x)^{-1/2}$. (And probably more posts about this can be found.)
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– Martin Sleziak
Jan 21 at 6:06
add a comment |
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Another way is
$$
eqalign{
& prodlimits_{k = 1}^n {{{2n - 2k + 1} over {2n - 2k + 2}}} = prodlimits_{k = 1}^n {{{left( {n - k} right) + 1/2} over {left( {n - k} right) + 1}}}
= prodlimits_{j = 0}^{n - 1} {{{j + 1/2} over {j + 1}}} = cr
& = {{left( {1/2} right)^{,overline {,n,} } } over {n!}}
= {{Gamma left( {n + 1/2} right)} over {Gamma left( {1/2} right)Gamma left( {n + 1} right)}} =
left( matrix{ n - 1/2 cr n cr} right)
= left( { - 1} right)^{,n} left( matrix{ - 1/2 cr n cr} right) cr}
$$
and then apply the binomial expansion.
answered Jan 20 at 23:47
G CabG Cab
19.8k31340
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Neat! thank you!
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– clathratus
Jan 20 at 23:57
add a comment |
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Another way is
$$
eqalign{
& prodlimits_{k = 1}^n {{{2n - 2k + 1} over {2n - 2k + 2}}} = prodlimits_{k = 1}^n {{{left( {n - k} right) + 1/2} over {left( {n - k} right) + 1}}}
= prodlimits_{j = 0}^{n - 1} {{{j + 1/2} over {j + 1}}} = cr
& = {{left( {1/2} right)^{,overline {,n,} } } over {n!}}
= {{Gamma left( {n + 1/2} right)} over {Gamma left( {1/2} right)Gamma left( {n + 1} right)}} =
left( matrix{ n - 1/2 cr n cr} right)
= left( { - 1} right)^{,n} left( matrix{ - 1/2 cr n cr} right) cr}
$$
and then apply the binomial expansion.
answered Jan 20 at 23:47
G CabG Cab
19.8k31340
$endgroup$
$begingroup$
Neat! thank you!
$endgroup$
– clathratus
Jan 20 at 23:57
add a comment |
$begingroup$
Another way is
$$
eqalign{
& prodlimits_{k = 1}^n {{{2n - 2k + 1} over {2n - 2k + 2}}} = prodlimits_{k = 1}^n {{{left( {n - k} right) + 1/2} over {left( {n - k} right) + 1}}}
= prodlimits_{j = 0}^{n - 1} {{{j + 1/2} over {j + 1}}} = cr
& = {{left( {1/2} right)^{,overline {,n,} } } over {n!}}
= {{Gamma left( {n + 1/2} right)} over {Gamma left( {1/2} right)Gamma left( {n + 1} right)}} =
left( matrix{ n - 1/2 cr n cr} right)
= left( { - 1} right)^{,n} left( matrix{ - 1/2 cr n cr} right) cr}
$$
and then apply the binomial expansion.
answered Jan 20 at 23:47
G CabG Cab
19.8k31340
$endgroup$
Another way is
$$
eqalign{
& prodlimits_{k = 1}^n {{{2n - 2k + 1} over {2n - 2k + 2}}} = prodlimits_{k = 1}^n {{{left( {n - k} right) + 1/2} over {left( {n - k} right) + 1}}}
= prodlimits_{j = 0}^{n - 1} {{{j + 1/2} over {j + 1}}} = cr
& = {{left( {1/2} right)^{,overline {,n,} } } over {n!}}
= {{Gamma left( {n + 1/2} right)} over {Gamma left( {1/2} right)Gamma left( {n + 1} right)}} =
left( matrix{ n - 1/2 cr n cr} right)
= left( { - 1} right)^{,n} left( matrix{ - 1/2 cr n cr} right) cr}
$$
and then apply the binomial expansion.
answered Jan 20 at 23:47
G CabG Cab
19.8k31340
answered Jan 20 at 23:47
G CabG Cab
19.8k31340
answered Jan 20 at 23:47
G CabG Cab
19.8k31340
answered Jan 20 at 23:47
G CabG Cab
19.8k31340
19.8k31340
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Neat! thank you!
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– clathratus
Jan 20 at 23:57
add a comment |
$begingroup$
Neat! thank you!
$endgroup$
– clathratus
Jan 20 at 23:57
$begingroup$
Neat! thank you!
$endgroup$
– clathratus
Jan 20 at 23:57
$begingroup$
Neat! thank you!
$endgroup$
– clathratus
Jan 20 at 23:57
add a comment |
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Satisfy yourself that $(0)_0=1$.
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– J.G.
Jan 20 at 22:30
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@J.G. I am now satisfied
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– clathratus
Jan 20 at 22:30
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In case of $n=0$, product in your sum is empty product, so you may regard it as 1. (Or just do the sum for $ngeq1$)
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– Seewoo Lee
Jan 20 at 22:31
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I don't think you need to include the $n=0$ case because the product is not defined for it.
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– Antinous
Jan 20 at 22:31
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@Antinous For $n=0$ the product is empty, and thus conventionally defined to equal $1$
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– Learner
Jan 20 at 22:59