Got stuck at a point -4

Given \(n\in\mathbb{Z}^+\) find condition for \(k\in\mathbb{Z}^+\) such that: \[\dfrac{2^{k+1}k!^2}{(2k+1)!}<\dfrac{1}{10^n}\] What I have done : \[\dfrac{2^{k+1}k!^2}{(2k+1)!}<\dfrac{1}{10^n}\] \[\Rightarrow 2^{k+n+1}5^n<\dfrac{(2k+1)!}{k!^2}\] \[=\dfrac{(2k+1)(2k)(2k-1)...(k+1)}{k!}\] \[\Rightarrow \ln a < \ln(\frac{(2k+1)(2k)(2k-1)...(k+1)}{k!})\space\space\space [\blue{a=2^{k+n+1}5^n}]\] \[=\sum_{n=k+1}^{2k+1}\ln n - \sum_{n=2}^{k}\ln n\] \[< \int_{k+1}^{2k+2}\ln x dx - \int_{1}^k \ln x dx\] \[=[(x-1)\ln x]_{k+1}^{2k+2}-[(x-1)\ln x]_{1}^{k}\] \[=(2k+1)\ln (2(k+1))-k\ln(k+1)-(k-1)\ln k\] \[=(2k+1)\ln(k+1)+(2k+1)\ln 2 -k\ln(k+1)-(k-1)\ln k\] \[=(k+1)\ln(k+1)+(2k+1)\ln 2-(k-1)\ln k\] \[\Rightarrow \ln a < (k+1)\ln(k+1)+(2k+1)\ln 2-(k-1)\ln k\] \[\Rightarrow a<\dfrac{(k+1)^{k+1}\times 2^{2k+1}}{k^{k-1}}\] \[\Rightarrow \dfrac{k^{k-1}}{(k+1)^{k+1}}<\dfrac{2^{k-n}}{5^n}\] Can anyone simplify it further? If yes then please comment !

Note by Zakir Husain
1 month, 1 week ago

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Actually, I feel the last equation is even more complex than initial equation we started with. Here is my method that gives an approximate condition for kk

2k+1k!2(2k+1)!<110n\frac{2^{k+1}k!^2}{(2k+1)!} \lt \frac1{10^n}

2k+1(2k+1)(2k)!k!2<110n\frac{2^{k+1}}{(2k+1)\frac{(2k)!}{k!^2}} \lt \frac1{10^n}

2k+1(2k+1)2k×(2k1)(2k3)1k!<110n\frac{2^{k+1}}{(2k+1)2^k × \frac{(2k-1)(2k-3)\ldots1}{k!}} \lt \frac1{10^n}

2(2+1k)(2+1k1)(2+1k2)(2+12)(2+11)<110n\frac{2}{(2+\frac1k)(2+\frac1{k-1})(2+\frac1{k-2})\ldots(2+\frac12)(2+\frac11)} \lt \frac1{10^n}

(2+11)(2+12)(2+1k1)(2+1k)>2×10n\displaystyle (2+\frac11)(2+\frac12)\ldots(2+\frac1{k-1})(2+\frac1k) \gt 2×10^n

3k>(2+11)(2+12)(2+1k1)(2+1k)>2×10n3^k \gt (2+\frac11)(2+\frac12)\ldots(2+\frac1{k-1})(2+\frac1k) \gt 2×10^n

3k>2×10n3^k \gt 2×10^n

k>2.096n+0.63k \gt 2.096n + 0.63

for better approximation, taking 11 term 2+112+\frac11 to right side

2.5k>(2+12)(2+13)(2+1k1)(2+1k)>23×10n2.5^k \gt (2+\frac12)(2+\frac13)\ldots(2+\frac1{k-1})(2+\frac1k) \gt \frac23×10^n

2.5k>2310n2.5^k \gt \frac23 10^n

k>2.51n0.44k \gt 2.51n - 0.44

For even better approximation, take one more term 2+122+\frac12 to right side

(73)k>(2+13)(2+14)(2+1k1)(2+1k)>27.5×10n(\frac73)^k \gt (2+\frac13)(2+\frac14)\ldots(2+\frac1{k-1})(2+\frac1k) \gt \frac2{7.5}×10^n

(73)k>41510n(\frac73)^k \gt \frac4{15} 10^n

k>2.717n1.56k \gt 2.717n - 1.56

For more better approximations, keep taking more and more terms to right side

Hope this helps. :)

@Zakir Husain

Aryan Sanghi - 1 month, 1 week ago

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Thanks for that

Zakir Husain - 1 month ago

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I think you did this for that π\pi's code, isn't it?

Aryan Sanghi - 1 month ago

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@Aryan Sanghi Yes I tried my program for f=1015000f=10^{15000} and it was not coming to an end, so I thought to approximate where it shall end.

Zakir Husain - 1 month ago

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Hmm... There is a new challenge now. We have to approximate j=1k(2+j1)\prod_{j=1}^k (2+j^{-1}), because my program have just crossed all the approximations we have done.

Zakir Husain - 1 month ago

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Not an approximation, but an alternate form.

2k+2(2k+32)!π(2k+3)k!\huge \frac{2^{k+2}(\frac{2k+3}{2})!}{\sqrt{π}(2k+3)k!}

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You could keep taking more and more terms to right side to get better approximations, as I said in last step. :)

Aryan Sanghi - 1 month ago

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