# 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
4 months, 2 weeks ago

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- 4 months, 2 weeks 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

- 4 months, 2 weeks ago

Thanks for that

- 4 months, 2 weeks ago

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

- 4 months, 2 weeks ago

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.

- 4 months, 2 weeks 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.

- 4 months, 2 weeks ago

Not an approximation, but an alternate form.

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

- 4 months, 2 weeks ago

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

- 4 months, 2 weeks ago