Using Wallis' product to derive $sqrtpi$












3












$begingroup$


Recall Wallis' product:
$$lim_{ntoinfty}Big(frac{2}{1}cdotfrac{2}{3}cdotfrac{4}{3}cdotfrac{4}{5}cdotfrac{6}{5}cdotsfrac{2n}{2n-1}cdotfrac{2n}{2n+1}Big)=frac{pi}{2}$$



We have to show that $$lim_{ntoinfty}frac{(n!)^22^{2n}}{(2n)!sqrt n}=sqrtpi$$



The hint I got was to use $$P_n=frac{(n!)^42^{4n}}{[(2n)!]^2(2n+1)}$$
which is just simply the inside of the limit in Wallis' product, multiplied and divided by $2cdot2cdot4cdot4cdots(2n)cdot(2n)$ alternatively. How do I use $P_n$ to derive $sqrtpi:$?






sequences-and-series analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    I think, you should need Stirling formula.
    $endgroup$
    – hamam_Abdallah
    Dec 8 '18 at 19:58










  • $begingroup$
    @hamam_Abdallah that comes later, so I can't use it.
    $endgroup$
    – Tomás Palamás
    Dec 8 '18 at 19:59
















3












$begingroup$


Recall Wallis' product:
$$lim_{ntoinfty}Big(frac{2}{1}cdotfrac{2}{3}cdotfrac{4}{3}cdotfrac{4}{5}cdotfrac{6}{5}cdotsfrac{2n}{2n-1}cdotfrac{2n}{2n+1}Big)=frac{pi}{2}$$



We have to show that $$lim_{ntoinfty}frac{(n!)^22^{2n}}{(2n)!sqrt n}=sqrtpi$$



The hint I got was to use $$P_n=frac{(n!)^42^{4n}}{[(2n)!]^2(2n+1)}$$
which is just simply the inside of the limit in Wallis' product, multiplied and divided by $2cdot2cdot4cdot4cdots(2n)cdot(2n)$ alternatively. How do I use $P_n$ to derive $sqrtpi:$?






sequences-and-series analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    I think, you should need Stirling formula.
    $endgroup$
    – hamam_Abdallah
    Dec 8 '18 at 19:58










  • $begingroup$
    @hamam_Abdallah that comes later, so I can't use it.
    $endgroup$
    – Tomás Palamás
    Dec 8 '18 at 19:59














3












3








3





$begingroup$


Recall Wallis' product:
$$lim_{ntoinfty}Big(frac{2}{1}cdotfrac{2}{3}cdotfrac{4}{3}cdotfrac{4}{5}cdotfrac{6}{5}cdotsfrac{2n}{2n-1}cdotfrac{2n}{2n+1}Big)=frac{pi}{2}$$



We have to show that $$lim_{ntoinfty}frac{(n!)^22^{2n}}{(2n)!sqrt n}=sqrtpi$$



The hint I got was to use $$P_n=frac{(n!)^42^{4n}}{[(2n)!]^2(2n+1)}$$
which is just simply the inside of the limit in Wallis' product, multiplied and divided by $2cdot2cdot4cdot4cdots(2n)cdot(2n)$ alternatively. How do I use $P_n$ to derive $sqrtpi:$?






sequences-and-series analysis







share|cite|improve this question









$endgroup$




Recall Wallis' product:
$$lim_{ntoinfty}Big(frac{2}{1}cdotfrac{2}{3}cdotfrac{4}{3}cdotfrac{4}{5}cdotfrac{6}{5}cdotsfrac{2n}{2n-1}cdotfrac{2n}{2n+1}Big)=frac{pi}{2}$$



We have to show that $$lim_{ntoinfty}frac{(n!)^22^{2n}}{(2n)!sqrt n}=sqrtpi$$



The hint I got was to use $$P_n=frac{(n!)^42^{4n}}{[(2n)!]^2(2n+1)}$$
which is just simply the inside of the limit in Wallis' product, multiplied and divided by $2cdot2cdot4cdot4cdots(2n)cdot(2n)$ alternatively. How do I use $P_n$ to derive $sqrtpi:$?






sequences-and-series analysis




sequences-and-series analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 8 '18 at 19:44









Tomás PalamásTomás Palamás

363211




363211












  • $begingroup$
    I think, you should need Stirling formula.
    $endgroup$
    – hamam_Abdallah
    Dec 8 '18 at 19:58










  • $begingroup$
    @hamam_Abdallah that comes later, so I can't use it.
    $endgroup$
    – Tomás Palamás
    Dec 8 '18 at 19:59


















  • $begingroup$
    I think, you should need Stirling formula.
    $endgroup$
    – hamam_Abdallah
    Dec 8 '18 at 19:58










  • $begingroup$
    @hamam_Abdallah that comes later, so I can't use it.
    $endgroup$
    – Tomás Palamás
    Dec 8 '18 at 19:59
















$begingroup$
I think, you should need Stirling formula.
$endgroup$
– hamam_Abdallah
Dec 8 '18 at 19:58




$begingroup$
I think, you should need Stirling formula.
$endgroup$
– hamam_Abdallah
Dec 8 '18 at 19:58












$begingroup$
@hamam_Abdallah that comes later, so I can't use it.
$endgroup$
– Tomás Palamás
Dec 8 '18 at 19:59




$begingroup$
@hamam_Abdallah that comes later, so I can't use it.
$endgroup$
– Tomás Palamás
Dec 8 '18 at 19:59










1 Answer
1






active

oldest

votes


















3












$begingroup$

As you noticed, $P_n to frac{pi}{2}$ since it's the inside of limit of the L.H.S of the Wallis Product Formula multiplied by $1$. Since continuous maps preserve limits, this implies $sqrt{2P_n} to sqrt{pi} $ and note that



$$lim_{n to infty}sqrt{2P_n} = lim_{n to infty}frac{sqrt{2}(n!)^2 2^{2n}}{(2n)!sqrt{2n+1 }} \ = lim_{n to infty} frac{(n!)^22^{2n}}{(2n)!sqrt{n+ frac{1}{sqrt{2}}}} \ = lim_{n to infty} frac{(n!)^22^{2n}}{(2n)! sqrt{n}}$$



Where the last equality was got by translation invariance.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I'm actually quite puzzled that you could have noticed that $P_n$ was the inside limit of the Wallis Product Formula multiplied by $1$ in a certain way but still be confused on how to answer this question.
    $endgroup$
    – Dionel Jaime
    Dec 8 '18 at 20:10










  • $begingroup$
    That was part of the hint, I just didn't want to write out the whole thing. I have two questions
    $endgroup$
    – Tomás Palamás
    Dec 8 '18 at 20:12










  • $begingroup$
    First, how did you multiply and take the square root knowing that they're limits?
    $endgroup$
    – Tomás Palamás
    Dec 8 '18 at 20:13










  • $begingroup$
    Second, how did you go from the first limit to the second?
    $endgroup$
    – Tomás Palamás
    Dec 8 '18 at 20:15










  • $begingroup$
    I edited the answer adding mildly more detail.
    $endgroup$
    – Dionel Jaime
    Dec 8 '18 at 20:24











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1 Answer
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3












$begingroup$

As you noticed, $P_n to frac{pi}{2}$ since it's the inside of limit of the L.H.S of the Wallis Product Formula multiplied by $1$. Since continuous maps preserve limits, this implies $sqrt{2P_n} to sqrt{pi} $ and note that



$$lim_{n to infty}sqrt{2P_n} = lim_{n to infty}frac{sqrt{2}(n!)^2 2^{2n}}{(2n)!sqrt{2n+1 }} \ = lim_{n to infty} frac{(n!)^22^{2n}}{(2n)!sqrt{n+ frac{1}{sqrt{2}}}} \ = lim_{n to infty} frac{(n!)^22^{2n}}{(2n)! sqrt{n}}$$



Where the last equality was got by translation invariance.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I'm actually quite puzzled that you could have noticed that $P_n$ was the inside limit of the Wallis Product Formula multiplied by $1$ in a certain way but still be confused on how to answer this question.
    $endgroup$
    – Dionel Jaime
    Dec 8 '18 at 20:10










  • $begingroup$
    That was part of the hint, I just didn't want to write out the whole thing. I have two questions
    $endgroup$
    – Tomás Palamás
    Dec 8 '18 at 20:12










  • $begingroup$
    First, how did you multiply and take the square root knowing that they're limits?
    $endgroup$
    – Tomás Palamás
    Dec 8 '18 at 20:13










  • $begingroup$
    Second, how did you go from the first limit to the second?
    $endgroup$
    – Tomás Palamás
    Dec 8 '18 at 20:15










  • $begingroup$
    I edited the answer adding mildly more detail.
    $endgroup$
    – Dionel Jaime
    Dec 8 '18 at 20:24
















3












$begingroup$

As you noticed, $P_n to frac{pi}{2}$ since it's the inside of limit of the L.H.S of the Wallis Product Formula multiplied by $1$. Since continuous maps preserve limits, this implies $sqrt{2P_n} to sqrt{pi} $ and note that



$$lim_{n to infty}sqrt{2P_n} = lim_{n to infty}frac{sqrt{2}(n!)^2 2^{2n}}{(2n)!sqrt{2n+1 }} \ = lim_{n to infty} frac{(n!)^22^{2n}}{(2n)!sqrt{n+ frac{1}{sqrt{2}}}} \ = lim_{n to infty} frac{(n!)^22^{2n}}{(2n)! sqrt{n}}$$



Where the last equality was got by translation invariance.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I'm actually quite puzzled that you could have noticed that $P_n$ was the inside limit of the Wallis Product Formula multiplied by $1$ in a certain way but still be confused on how to answer this question.
    $endgroup$
    – Dionel Jaime
    Dec 8 '18 at 20:10










  • $begingroup$
    That was part of the hint, I just didn't want to write out the whole thing. I have two questions
    $endgroup$
    – Tomás Palamás
    Dec 8 '18 at 20:12










  • $begingroup$
    First, how did you multiply and take the square root knowing that they're limits?
    $endgroup$
    – Tomás Palamás
    Dec 8 '18 at 20:13










  • $begingroup$
    Second, how did you go from the first limit to the second?
    $endgroup$
    – Tomás Palamás
    Dec 8 '18 at 20:15










  • $begingroup$
    I edited the answer adding mildly more detail.
    $endgroup$
    – Dionel Jaime
    Dec 8 '18 at 20:24














3












3








3





$begingroup$

As you noticed, $P_n to frac{pi}{2}$ since it's the inside of limit of the L.H.S of the Wallis Product Formula multiplied by $1$. Since continuous maps preserve limits, this implies $sqrt{2P_n} to sqrt{pi} $ and note that



$$lim_{n to infty}sqrt{2P_n} = lim_{n to infty}frac{sqrt{2}(n!)^2 2^{2n}}{(2n)!sqrt{2n+1 }} \ = lim_{n to infty} frac{(n!)^22^{2n}}{(2n)!sqrt{n+ frac{1}{sqrt{2}}}} \ = lim_{n to infty} frac{(n!)^22^{2n}}{(2n)! sqrt{n}}$$



Where the last equality was got by translation invariance.






share|cite|improve this answer











$endgroup$



As you noticed, $P_n to frac{pi}{2}$ since it's the inside of limit of the L.H.S of the Wallis Product Formula multiplied by $1$. Since continuous maps preserve limits, this implies $sqrt{2P_n} to sqrt{pi} $ and note that



$$lim_{n to infty}sqrt{2P_n} = lim_{n to infty}frac{sqrt{2}(n!)^2 2^{2n}}{(2n)!sqrt{2n+1 }} \ = lim_{n to infty} frac{(n!)^22^{2n}}{(2n)!sqrt{n+ frac{1}{sqrt{2}}}} \ = lim_{n to infty} frac{(n!)^22^{2n}}{(2n)! sqrt{n}}$$



Where the last equality was got by translation invariance.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 9 '18 at 3:47

























answered Dec 8 '18 at 20:03









Dionel JaimeDionel Jaime

1,743514




1,743514












  • $begingroup$
    I'm actually quite puzzled that you could have noticed that $P_n$ was the inside limit of the Wallis Product Formula multiplied by $1$ in a certain way but still be confused on how to answer this question.
    $endgroup$
    – Dionel Jaime
    Dec 8 '18 at 20:10










  • $begingroup$
    That was part of the hint, I just didn't want to write out the whole thing. I have two questions
    $endgroup$
    – Tomás Palamás
    Dec 8 '18 at 20:12










  • $begingroup$
    First, how did you multiply and take the square root knowing that they're limits?
    $endgroup$
    – Tomás Palamás
    Dec 8 '18 at 20:13










  • $begingroup$
    Second, how did you go from the first limit to the second?
    $endgroup$
    – Tomás Palamás
    Dec 8 '18 at 20:15










  • $begingroup$
    I edited the answer adding mildly more detail.
    $endgroup$
    – Dionel Jaime
    Dec 8 '18 at 20:24


















  • $begingroup$
    I'm actually quite puzzled that you could have noticed that $P_n$ was the inside limit of the Wallis Product Formula multiplied by $1$ in a certain way but still be confused on how to answer this question.
    $endgroup$
    – Dionel Jaime
    Dec 8 '18 at 20:10










  • $begingroup$
    That was part of the hint, I just didn't want to write out the whole thing. I have two questions
    $endgroup$
    – Tomás Palamás
    Dec 8 '18 at 20:12










  • $begingroup$
    First, how did you multiply and take the square root knowing that they're limits?
    $endgroup$
    – Tomás Palamás
    Dec 8 '18 at 20:13










  • $begingroup$
    Second, how did you go from the first limit to the second?
    $endgroup$
    – Tomás Palamás
    Dec 8 '18 at 20:15










  • $begingroup$
    I edited the answer adding mildly more detail.
    $endgroup$
    – Dionel Jaime
    Dec 8 '18 at 20:24
















$begingroup$
I'm actually quite puzzled that you could have noticed that $P_n$ was the inside limit of the Wallis Product Formula multiplied by $1$ in a certain way but still be confused on how to answer this question.
$endgroup$
– Dionel Jaime
Dec 8 '18 at 20:10




$begingroup$
I'm actually quite puzzled that you could have noticed that $P_n$ was the inside limit of the Wallis Product Formula multiplied by $1$ in a certain way but still be confused on how to answer this question.
$endgroup$
– Dionel Jaime
Dec 8 '18 at 20:10












$begingroup$
That was part of the hint, I just didn't want to write out the whole thing. I have two questions
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:12




$begingroup$
That was part of the hint, I just didn't want to write out the whole thing. I have two questions
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:12












$begingroup$
First, how did you multiply and take the square root knowing that they're limits?
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:13




$begingroup$
First, how did you multiply and take the square root knowing that they're limits?
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:13












$begingroup$
Second, how did you go from the first limit to the second?
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:15




$begingroup$
Second, how did you go from the first limit to the second?
$endgroup$
– Tomás Palamás
Dec 8 '18 at 20:15












$begingroup$
I edited the answer adding mildly more detail.
$endgroup$
– Dionel Jaime
Dec 8 '18 at 20:24




$begingroup$
I edited the answer adding mildly more detail.
$endgroup$
– Dionel Jaime
Dec 8 '18 at 20:24


















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