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Prove that $1+frac{1}{8n}sqrt{pi_1n}


Evaluating $sqrt{1 + sqrt{2 + sqrt{4 + sqrt{8 + ldots}}}}$Prove the inequality $,frac{1}{sqrt{1}+ sqrt{3}} +frac{1}{sqrt{5}+ sqrt{7} }+ldots+frac{1}{sqrt{9997}+sqrt{9999}}gt 24$Prove or disprove: $sqrt{2sqrt{3sqrt{4sqrt{ldotssqrt{n}}}}}<3$Calculate $sqrt{frac{1}{2}} times sqrt{frac{1}{2} + frac{1}{2}sqrt{frac{1}{2}}} times ldots $Prove that $sqrt{k}+frac{1}{sqrt{k}+sqrt{k+1}}=sqrt{k+1}$About $sqrt[2]{1+sqrt[3]{1+sqrt[4]{1+sqrt[5]{1+ldots}}}}$Prove that $frac{1}{1 +sqrt{2}}+frac{1}{sqrt{2} +sqrt{3}}+­…+frac{1}{sqrt{99} +sqrt{100}}=9$.Prove that $frac{1}{sqrt{nx+1}}+frac{1}{sqrt{nx+2}}+ldots+frac{1}{sqrt{nx+n}}=sqrt n$ has a unique positive solutionProve that $ln2<frac{1}{sqrt[3]3}$$ln n+sqrt{frac12}+sqrt{frac23}+ldots +sqrt{frac{n-1}{n}}<sqrt2+sqrt{frac32}+sqrt{frac43}+ldots +sqrt{frac{n}{n-1}}$













2












$begingroup$


Within the confines of the O-level syllabus, prove that:



$$1+dfrac{1}{8n}sqrt{pi_1n}<dfrac{2.4.6ldots(2n-2)(2n)}{1.3.5ldots(2n-3)(2n-1)}<left(1+dfrac{1}{8n}+dfrac{1}{128n^{2}}right)sqrt{pi_2n}$$



for all positive integer $n$, where $pi_1=3.141$ and $pi_2=3.142$



Using A-level mathematics, show that the result remains true with $pi_1=pi_2=pi$ where $pi$ is the familiar constant associated with a circle.



This is a question found in Hammersley's "On the enfeeblement of mathematical skills by "Modern Mathematics" and by similar soft intellectual trash in schools and universities"



When I first look at this problem, I wonder if I can prove it via Wallis'product?
$$prod_{i=1}^{infty}dfrac{2n}{2n-1}.dfrac{2n}{2n+1}=dfrac{pi}{2}$$






calculus algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why downvote this question?
    $endgroup$
    – James Warthington
    yesterday










  • $begingroup$
    I doubt whether Wallis's product has ever featured in an A-level (let alone O-level) syllabus.
    $endgroup$
    – Lord Shark the Unknown
    yesterday










  • $begingroup$
    Thanks, I am not familiar with British school curriculum. The author is an Englishman anyway.
    $endgroup$
    – James Warthington
    yesterday






  • 1




    $begingroup$
    You wrote "where $pi_1=3.141$ and $pi_1=3.142$". Did you mean the second variable to be $pi_2$?
    $endgroup$
    – John Omielan
    yesterday










  • $begingroup$
    I don't understand your quesion. $pi_2$ is the second number. Both $pi_1$ and $pi_2$ are numbers that sandwich $pi$.
    $endgroup$
    – James Warthington
    yesterday


















2












$begingroup$


Within the confines of the O-level syllabus, prove that:



$$1+dfrac{1}{8n}sqrt{pi_1n}<dfrac{2.4.6ldots(2n-2)(2n)}{1.3.5ldots(2n-3)(2n-1)}<left(1+dfrac{1}{8n}+dfrac{1}{128n^{2}}right)sqrt{pi_2n}$$



for all positive integer $n$, where $pi_1=3.141$ and $pi_2=3.142$



Using A-level mathematics, show that the result remains true with $pi_1=pi_2=pi$ where $pi$ is the familiar constant associated with a circle.



This is a question found in Hammersley's "On the enfeeblement of mathematical skills by "Modern Mathematics" and by similar soft intellectual trash in schools and universities"



When I first look at this problem, I wonder if I can prove it via Wallis'product?
$$prod_{i=1}^{infty}dfrac{2n}{2n-1}.dfrac{2n}{2n+1}=dfrac{pi}{2}$$






calculus algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why downvote this question?
    $endgroup$
    – James Warthington
    yesterday










  • $begingroup$
    I doubt whether Wallis's product has ever featured in an A-level (let alone O-level) syllabus.
    $endgroup$
    – Lord Shark the Unknown
    yesterday










  • $begingroup$
    Thanks, I am not familiar with British school curriculum. The author is an Englishman anyway.
    $endgroup$
    – James Warthington
    yesterday






  • 1




    $begingroup$
    You wrote "where $pi_1=3.141$ and $pi_1=3.142$". Did you mean the second variable to be $pi_2$?
    $endgroup$
    – John Omielan
    yesterday










  • $begingroup$
    I don't understand your quesion. $pi_2$ is the second number. Both $pi_1$ and $pi_2$ are numbers that sandwich $pi$.
    $endgroup$
    – James Warthington
    yesterday
















2












2








2


1



$begingroup$


Within the confines of the O-level syllabus, prove that:



$$1+dfrac{1}{8n}sqrt{pi_1n}<dfrac{2.4.6ldots(2n-2)(2n)}{1.3.5ldots(2n-3)(2n-1)}<left(1+dfrac{1}{8n}+dfrac{1}{128n^{2}}right)sqrt{pi_2n}$$



for all positive integer $n$, where $pi_1=3.141$ and $pi_2=3.142$



Using A-level mathematics, show that the result remains true with $pi_1=pi_2=pi$ where $pi$ is the familiar constant associated with a circle.



This is a question found in Hammersley's "On the enfeeblement of mathematical skills by "Modern Mathematics" and by similar soft intellectual trash in schools and universities"



When I first look at this problem, I wonder if I can prove it via Wallis'product?
$$prod_{i=1}^{infty}dfrac{2n}{2n-1}.dfrac{2n}{2n+1}=dfrac{pi}{2}$$






calculus algebra-precalculus







share|cite|improve this question











$endgroup$




Within the confines of the O-level syllabus, prove that:



$$1+dfrac{1}{8n}sqrt{pi_1n}<dfrac{2.4.6ldots(2n-2)(2n)}{1.3.5ldots(2n-3)(2n-1)}<left(1+dfrac{1}{8n}+dfrac{1}{128n^{2}}right)sqrt{pi_2n}$$



for all positive integer $n$, where $pi_1=3.141$ and $pi_2=3.142$



Using A-level mathematics, show that the result remains true with $pi_1=pi_2=pi$ where $pi$ is the familiar constant associated with a circle.



This is a question found in Hammersley's "On the enfeeblement of mathematical skills by "Modern Mathematics" and by similar soft intellectual trash in schools and universities"



When I first look at this problem, I wonder if I can prove it via Wallis'product?
$$prod_{i=1}^{infty}dfrac{2n}{2n-1}.dfrac{2n}{2n+1}=dfrac{pi}{2}$$






calculus algebra-precalculus




calculus algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









StubbornAtom

6,07311239




6,07311239










asked yesterday









James WarthingtonJames Warthington

47729




47729












  • $begingroup$
    Why downvote this question?
    $endgroup$
    – James Warthington
    yesterday










  • $begingroup$
    I doubt whether Wallis's product has ever featured in an A-level (let alone O-level) syllabus.
    $endgroup$
    – Lord Shark the Unknown
    yesterday










  • $begingroup$
    Thanks, I am not familiar with British school curriculum. The author is an Englishman anyway.
    $endgroup$
    – James Warthington
    yesterday






  • 1




    $begingroup$
    You wrote "where $pi_1=3.141$ and $pi_1=3.142$". Did you mean the second variable to be $pi_2$?
    $endgroup$
    – John Omielan
    yesterday










  • $begingroup$
    I don't understand your quesion. $pi_2$ is the second number. Both $pi_1$ and $pi_2$ are numbers that sandwich $pi$.
    $endgroup$
    – James Warthington
    yesterday




















  • $begingroup$
    Why downvote this question?
    $endgroup$
    – James Warthington
    yesterday










  • $begingroup$
    I doubt whether Wallis's product has ever featured in an A-level (let alone O-level) syllabus.
    $endgroup$
    – Lord Shark the Unknown
    yesterday










  • $begingroup$
    Thanks, I am not familiar with British school curriculum. The author is an Englishman anyway.
    $endgroup$
    – James Warthington
    yesterday






  • 1




    $begingroup$
    You wrote "where $pi_1=3.141$ and $pi_1=3.142$". Did you mean the second variable to be $pi_2$?
    $endgroup$
    – John Omielan
    yesterday










  • $begingroup$
    I don't understand your quesion. $pi_2$ is the second number. Both $pi_1$ and $pi_2$ are numbers that sandwich $pi$.
    $endgroup$
    – James Warthington
    yesterday


















$begingroup$
Why downvote this question?
$endgroup$
– James Warthington
yesterday




$begingroup$
Why downvote this question?
$endgroup$
– James Warthington
yesterday












$begingroup$
I doubt whether Wallis's product has ever featured in an A-level (let alone O-level) syllabus.
$endgroup$
– Lord Shark the Unknown
yesterday




$begingroup$
I doubt whether Wallis's product has ever featured in an A-level (let alone O-level) syllabus.
$endgroup$
– Lord Shark the Unknown
yesterday












$begingroup$
Thanks, I am not familiar with British school curriculum. The author is an Englishman anyway.
$endgroup$
– James Warthington
yesterday




$begingroup$
Thanks, I am not familiar with British school curriculum. The author is an Englishman anyway.
$endgroup$
– James Warthington
yesterday




1




1




$begingroup$
You wrote "where $pi_1=3.141$ and $pi_1=3.142$". Did you mean the second variable to be $pi_2$?
$endgroup$
– John Omielan
yesterday




$begingroup$
You wrote "where $pi_1=3.141$ and $pi_1=3.142$". Did you mean the second variable to be $pi_2$?
$endgroup$
– John Omielan
yesterday












$begingroup$
I don't understand your quesion. $pi_2$ is the second number. Both $pi_1$ and $pi_2$ are numbers that sandwich $pi$.
$endgroup$
– James Warthington
yesterday






$begingroup$
I don't understand your quesion. $pi_2$ is the second number. Both $pi_1$ and $pi_2$ are numbers that sandwich $pi$.
$endgroup$
– James Warthington
yesterday












1 Answer
1






active

oldest

votes


















0












$begingroup$

$$
dfrac{2.4.6ldots(2n-2)(2n)}{1.3.5ldots(2n-3)(2n-1)} = dfrac{2n!}{(2n-1)!} = 2n \ therefore ;; 1+dfrac{1}{8n}sqrt{pi_1n}<dfrac{2.4.6ldots(2n-2)(2n)}{1.3.5ldots(2n-3)(2n-1)}<left(1+dfrac{1}{8n}+dfrac{1}{128n^{2}}right)sqrt{pi_2n}
\ = 1+dfrac{1}{8n}sqrt{pi_1n}<2n<left(1+dfrac{1}{8n}+dfrac{1}{128n^{2}}right)sqrt{pi_2n}
$$



Considering the case of $$pi_1 = pi_2 = pi$$ since this generality can be used for the other case too.



Splitting the inequalities into two,



begin{align}
I_1:
1+dfrac{1}{8n}sqrt{pi n}<2n \
1+dfrac{1}{8sqrt n}sqrt{pi}<2n \
dfrac{1}{8sqrt n}sqrt{pi}<2n - 1 \
dfrac{1}{8}sqrt{pi}<(2n - 1)cdotsqrt n tag{1}label{1}
end{align}
Since n is a positive integer, minimum value of n is 1, which when substituted in eqref{1} gives $$ 0.2215 < 1 $$ which is true.



begin{align}
I_2:
2n<left(1+dfrac{1}{8n}+dfrac{1}{128n^{2}}right)sqrt{pi n}\
dfrac{2n}{sqrt{pi n}}<left(1+dfrac{1}{8n}+dfrac{1}{128n^{2}}right) \
dfrac{2n}{sqrt{pi n}} - 1<left(dfrac{1}{8n}+dfrac{1}{128n^{2}}right) \
dfrac{2n}{sqrt{pi n}} - 1<dfrac{1}{8n}left(1+dfrac{1}{16n}right)\
dfrac{2n}{sqrt{pi n}} - 1<dfrac{1}{8n}left(dfrac{16n+1}{16n}right)\
(128n^{2})cdotdfrac{2sqrt n}{sqrt{pi}} - 1< 16n+1\
dfrac{256n^{frac{5}{2}}}{sqrt pi} - 128n^{2}<16n+1\
dfrac{256}{sqrt pi} < ( 128n^2 + 16n + 1 )cdot n^{frac{2}{5}} tag{2}label{2}
end{align}



Since n is a positive integer, minimum value of n is 1, which when substituted in eqref{2} gives $$ 144.43 < 145 $$ which is true.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This [ $dfrac{2.4.6ldots(2n-2)(2n)}{1.3.5ldots(2n-3)(2n-1)} = dfrac{2n!}{(2n-1)!} = 2n$ ] is completely false (the numerator is the product of just even numbers up to $2n$ and not all the numbers as you assume). For example for $n=2$ we have $frac{2cdot 4}{1cdot 3} = frac{8}{3}$ which is not equal to $2n = 4$.
    $endgroup$
    – Winther
    yesterday












  • $begingroup$
    @Winther Oh right, I assumed it as n in my head, ugh all that work gone in trash.
    $endgroup$
    – Ashok
    yesterday













Your Answer





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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









0












$begingroup$

$$
dfrac{2.4.6ldots(2n-2)(2n)}{1.3.5ldots(2n-3)(2n-1)} = dfrac{2n!}{(2n-1)!} = 2n \ therefore ;; 1+dfrac{1}{8n}sqrt{pi_1n}<dfrac{2.4.6ldots(2n-2)(2n)}{1.3.5ldots(2n-3)(2n-1)}<left(1+dfrac{1}{8n}+dfrac{1}{128n^{2}}right)sqrt{pi_2n}
\ = 1+dfrac{1}{8n}sqrt{pi_1n}<2n<left(1+dfrac{1}{8n}+dfrac{1}{128n^{2}}right)sqrt{pi_2n}
$$



Considering the case of $$pi_1 = pi_2 = pi$$ since this generality can be used for the other case too.



Splitting the inequalities into two,



begin{align}
I_1:
1+dfrac{1}{8n}sqrt{pi n}<2n \
1+dfrac{1}{8sqrt n}sqrt{pi}<2n \
dfrac{1}{8sqrt n}sqrt{pi}<2n - 1 \
dfrac{1}{8}sqrt{pi}<(2n - 1)cdotsqrt n tag{1}label{1}
end{align}
Since n is a positive integer, minimum value of n is 1, which when substituted in eqref{1} gives $$ 0.2215 < 1 $$ which is true.



begin{align}
I_2:
2n<left(1+dfrac{1}{8n}+dfrac{1}{128n^{2}}right)sqrt{pi n}\
dfrac{2n}{sqrt{pi n}}<left(1+dfrac{1}{8n}+dfrac{1}{128n^{2}}right) \
dfrac{2n}{sqrt{pi n}} - 1<left(dfrac{1}{8n}+dfrac{1}{128n^{2}}right) \
dfrac{2n}{sqrt{pi n}} - 1<dfrac{1}{8n}left(1+dfrac{1}{16n}right)\
dfrac{2n}{sqrt{pi n}} - 1<dfrac{1}{8n}left(dfrac{16n+1}{16n}right)\
(128n^{2})cdotdfrac{2sqrt n}{sqrt{pi}} - 1< 16n+1\
dfrac{256n^{frac{5}{2}}}{sqrt pi} - 128n^{2}<16n+1\
dfrac{256}{sqrt pi} < ( 128n^2 + 16n + 1 )cdot n^{frac{2}{5}} tag{2}label{2}
end{align}



Since n is a positive integer, minimum value of n is 1, which when substituted in eqref{2} gives $$ 144.43 < 145 $$ which is true.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This [ $dfrac{2.4.6ldots(2n-2)(2n)}{1.3.5ldots(2n-3)(2n-1)} = dfrac{2n!}{(2n-1)!} = 2n$ ] is completely false (the numerator is the product of just even numbers up to $2n$ and not all the numbers as you assume). For example for $n=2$ we have $frac{2cdot 4}{1cdot 3} = frac{8}{3}$ which is not equal to $2n = 4$.
    $endgroup$
    – Winther
    yesterday












  • $begingroup$
    @Winther Oh right, I assumed it as n in my head, ugh all that work gone in trash.
    $endgroup$
    – Ashok
    yesterday


















0












$begingroup$

$$
dfrac{2.4.6ldots(2n-2)(2n)}{1.3.5ldots(2n-3)(2n-1)} = dfrac{2n!}{(2n-1)!} = 2n \ therefore ;; 1+dfrac{1}{8n}sqrt{pi_1n}<dfrac{2.4.6ldots(2n-2)(2n)}{1.3.5ldots(2n-3)(2n-1)}<left(1+dfrac{1}{8n}+dfrac{1}{128n^{2}}right)sqrt{pi_2n}
\ = 1+dfrac{1}{8n}sqrt{pi_1n}<2n<left(1+dfrac{1}{8n}+dfrac{1}{128n^{2}}right)sqrt{pi_2n}
$$



Considering the case of $$pi_1 = pi_2 = pi$$ since this generality can be used for the other case too.



Splitting the inequalities into two,



begin{align}
I_1:
1+dfrac{1}{8n}sqrt{pi n}<2n \
1+dfrac{1}{8sqrt n}sqrt{pi}<2n \
dfrac{1}{8sqrt n}sqrt{pi}<2n - 1 \
dfrac{1}{8}sqrt{pi}<(2n - 1)cdotsqrt n tag{1}label{1}
end{align}
Since n is a positive integer, minimum value of n is 1, which when substituted in eqref{1} gives $$ 0.2215 < 1 $$ which is true.



begin{align}
I_2:
2n<left(1+dfrac{1}{8n}+dfrac{1}{128n^{2}}right)sqrt{pi n}\
dfrac{2n}{sqrt{pi n}}<left(1+dfrac{1}{8n}+dfrac{1}{128n^{2}}right) \
dfrac{2n}{sqrt{pi n}} - 1<left(dfrac{1}{8n}+dfrac{1}{128n^{2}}right) \
dfrac{2n}{sqrt{pi n}} - 1<dfrac{1}{8n}left(1+dfrac{1}{16n}right)\
dfrac{2n}{sqrt{pi n}} - 1<dfrac{1}{8n}left(dfrac{16n+1}{16n}right)\
(128n^{2})cdotdfrac{2sqrt n}{sqrt{pi}} - 1< 16n+1\
dfrac{256n^{frac{5}{2}}}{sqrt pi} - 128n^{2}<16n+1\
dfrac{256}{sqrt pi} < ( 128n^2 + 16n + 1 )cdot n^{frac{2}{5}} tag{2}label{2}
end{align}



Since n is a positive integer, minimum value of n is 1, which when substituted in eqref{2} gives $$ 144.43 < 145 $$ which is true.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This [ $dfrac{2.4.6ldots(2n-2)(2n)}{1.3.5ldots(2n-3)(2n-1)} = dfrac{2n!}{(2n-1)!} = 2n$ ] is completely false (the numerator is the product of just even numbers up to $2n$ and not all the numbers as you assume). For example for $n=2$ we have $frac{2cdot 4}{1cdot 3} = frac{8}{3}$ which is not equal to $2n = 4$.
    $endgroup$
    – Winther
    yesterday












  • $begingroup$
    @Winther Oh right, I assumed it as n in my head, ugh all that work gone in trash.
    $endgroup$
    – Ashok
    yesterday
















0












0








0





$begingroup$

$$
dfrac{2.4.6ldots(2n-2)(2n)}{1.3.5ldots(2n-3)(2n-1)} = dfrac{2n!}{(2n-1)!} = 2n \ therefore ;; 1+dfrac{1}{8n}sqrt{pi_1n}<dfrac{2.4.6ldots(2n-2)(2n)}{1.3.5ldots(2n-3)(2n-1)}<left(1+dfrac{1}{8n}+dfrac{1}{128n^{2}}right)sqrt{pi_2n}
\ = 1+dfrac{1}{8n}sqrt{pi_1n}<2n<left(1+dfrac{1}{8n}+dfrac{1}{128n^{2}}right)sqrt{pi_2n}
$$



Considering the case of $$pi_1 = pi_2 = pi$$ since this generality can be used for the other case too.



Splitting the inequalities into two,



begin{align}
I_1:
1+dfrac{1}{8n}sqrt{pi n}<2n \
1+dfrac{1}{8sqrt n}sqrt{pi}<2n \
dfrac{1}{8sqrt n}sqrt{pi}<2n - 1 \
dfrac{1}{8}sqrt{pi}<(2n - 1)cdotsqrt n tag{1}label{1}
end{align}
Since n is a positive integer, minimum value of n is 1, which when substituted in eqref{1} gives $$ 0.2215 < 1 $$ which is true.



begin{align}
I_2:
2n<left(1+dfrac{1}{8n}+dfrac{1}{128n^{2}}right)sqrt{pi n}\
dfrac{2n}{sqrt{pi n}}<left(1+dfrac{1}{8n}+dfrac{1}{128n^{2}}right) \
dfrac{2n}{sqrt{pi n}} - 1<left(dfrac{1}{8n}+dfrac{1}{128n^{2}}right) \
dfrac{2n}{sqrt{pi n}} - 1<dfrac{1}{8n}left(1+dfrac{1}{16n}right)\
dfrac{2n}{sqrt{pi n}} - 1<dfrac{1}{8n}left(dfrac{16n+1}{16n}right)\
(128n^{2})cdotdfrac{2sqrt n}{sqrt{pi}} - 1< 16n+1\
dfrac{256n^{frac{5}{2}}}{sqrt pi} - 128n^{2}<16n+1\
dfrac{256}{sqrt pi} < ( 128n^2 + 16n + 1 )cdot n^{frac{2}{5}} tag{2}label{2}
end{align}



Since n is a positive integer, minimum value of n is 1, which when substituted in eqref{2} gives $$ 144.43 < 145 $$ which is true.






share|cite|improve this answer









$endgroup$



$$
dfrac{2.4.6ldots(2n-2)(2n)}{1.3.5ldots(2n-3)(2n-1)} = dfrac{2n!}{(2n-1)!} = 2n \ therefore ;; 1+dfrac{1}{8n}sqrt{pi_1n}<dfrac{2.4.6ldots(2n-2)(2n)}{1.3.5ldots(2n-3)(2n-1)}<left(1+dfrac{1}{8n}+dfrac{1}{128n^{2}}right)sqrt{pi_2n}
\ = 1+dfrac{1}{8n}sqrt{pi_1n}<2n<left(1+dfrac{1}{8n}+dfrac{1}{128n^{2}}right)sqrt{pi_2n}
$$



Considering the case of $$pi_1 = pi_2 = pi$$ since this generality can be used for the other case too.



Splitting the inequalities into two,



begin{align}
I_1:
1+dfrac{1}{8n}sqrt{pi n}<2n \
1+dfrac{1}{8sqrt n}sqrt{pi}<2n \
dfrac{1}{8sqrt n}sqrt{pi}<2n - 1 \
dfrac{1}{8}sqrt{pi}<(2n - 1)cdotsqrt n tag{1}label{1}
end{align}
Since n is a positive integer, minimum value of n is 1, which when substituted in eqref{1} gives $$ 0.2215 < 1 $$ which is true.



begin{align}
I_2:
2n<left(1+dfrac{1}{8n}+dfrac{1}{128n^{2}}right)sqrt{pi n}\
dfrac{2n}{sqrt{pi n}}<left(1+dfrac{1}{8n}+dfrac{1}{128n^{2}}right) \
dfrac{2n}{sqrt{pi n}} - 1<left(dfrac{1}{8n}+dfrac{1}{128n^{2}}right) \
dfrac{2n}{sqrt{pi n}} - 1<dfrac{1}{8n}left(1+dfrac{1}{16n}right)\
dfrac{2n}{sqrt{pi n}} - 1<dfrac{1}{8n}left(dfrac{16n+1}{16n}right)\
(128n^{2})cdotdfrac{2sqrt n}{sqrt{pi}} - 1< 16n+1\
dfrac{256n^{frac{5}{2}}}{sqrt pi} - 128n^{2}<16n+1\
dfrac{256}{sqrt pi} < ( 128n^2 + 16n + 1 )cdot n^{frac{2}{5}} tag{2}label{2}
end{align}



Since n is a positive integer, minimum value of n is 1, which when substituted in eqref{2} gives $$ 144.43 < 145 $$ which is true.







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answered yesterday









AshokAshok

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  • $begingroup$
    This [ $dfrac{2.4.6ldots(2n-2)(2n)}{1.3.5ldots(2n-3)(2n-1)} = dfrac{2n!}{(2n-1)!} = 2n$ ] is completely false (the numerator is the product of just even numbers up to $2n$ and not all the numbers as you assume). For example for $n=2$ we have $frac{2cdot 4}{1cdot 3} = frac{8}{3}$ which is not equal to $2n = 4$.
    $endgroup$
    – Winther
    yesterday












  • $begingroup$
    @Winther Oh right, I assumed it as n in my head, ugh all that work gone in trash.
    $endgroup$
    – Ashok
    yesterday




















  • $begingroup$
    This [ $dfrac{2.4.6ldots(2n-2)(2n)}{1.3.5ldots(2n-3)(2n-1)} = dfrac{2n!}{(2n-1)!} = 2n$ ] is completely false (the numerator is the product of just even numbers up to $2n$ and not all the numbers as you assume). For example for $n=2$ we have $frac{2cdot 4}{1cdot 3} = frac{8}{3}$ which is not equal to $2n = 4$.
    $endgroup$
    – Winther
    yesterday












  • $begingroup$
    @Winther Oh right, I assumed it as n in my head, ugh all that work gone in trash.
    $endgroup$
    – Ashok
    yesterday


















$begingroup$
This [ $dfrac{2.4.6ldots(2n-2)(2n)}{1.3.5ldots(2n-3)(2n-1)} = dfrac{2n!}{(2n-1)!} = 2n$ ] is completely false (the numerator is the product of just even numbers up to $2n$ and not all the numbers as you assume). For example for $n=2$ we have $frac{2cdot 4}{1cdot 3} = frac{8}{3}$ which is not equal to $2n = 4$.
$endgroup$
– Winther
yesterday






$begingroup$
This [ $dfrac{2.4.6ldots(2n-2)(2n)}{1.3.5ldots(2n-3)(2n-1)} = dfrac{2n!}{(2n-1)!} = 2n$ ] is completely false (the numerator is the product of just even numbers up to $2n$ and not all the numbers as you assume). For example for $n=2$ we have $frac{2cdot 4}{1cdot 3} = frac{8}{3}$ which is not equal to $2n = 4$.
$endgroup$
– Winther
yesterday














$begingroup$
@Winther Oh right, I assumed it as n in my head, ugh all that work gone in trash.
$endgroup$
– Ashok
yesterday






$begingroup$
@Winther Oh right, I assumed it as n in my head, ugh all that work gone in trash.
$endgroup$
– Ashok
yesterday




















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