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 !

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Comments

@Aryan Sanghi @Siddharth Chakravarty @Yajat Shamji @Mahdi Raza @Kumudesh Ghosh @Marvin Kalngan

Zakir Husain - 3 months ago

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 $k$

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

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

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

$\frac{2}{(2+\frac1k)(2+\frac1{k-1})(2+\frac1{k-2})\ldots(2+\frac12)(2+\frac11)} \lt \frac1{10^n}$

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

$3^k \gt (2+\frac11)(2+\frac12)\ldots(2+\frac1{k-1})(2+\frac1k) \gt 2×10^n$

$3^k \gt 2×10^n$

$k \gt 2.096n + 0.63$

for better approximation, taking $1$ term $2+\frac11$ to right side

$2.5^k \gt (2+\frac12)(2+\frac13)\ldots(2+\frac1{k-1})(2+\frac1k) \gt \frac23×10^n$

$2.5^k \gt \frac23 10^n$

$k \gt 2.51n - 0.44$

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

$(\frac73)^k \gt (2+\frac13)(2+\frac14)\ldots(2+\frac1{k-1})(2+\frac1k) \gt \frac2{7.5}×10^n$

$(\frac73)^k \gt \frac4{15} 10^n$

$k \gt 2.717n - 1.56$

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

Hope this helps. :)

@Zakir Husain

Aryan Sanghi - 3 months ago

Thanks for that

Zakir Husain - 3 months ago

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

Aryan Sanghi - 3 months ago

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

Zakir Husain - 3 months ago

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

Zakir Husain - 3 months ago

Not an approximation, but an alternate form.

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

Siddharth Chakravarty - 3 months ago

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

Aryan Sanghi - 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$