Find the integral part of $S$.

Question

It is given that $S=\sqrt{2012\sqrt{2013\sqrt{2014\sqrt{\cdots \sqrt{(2012^2-2)\sqrt{(2012^2-1)\sqrt{2012^2}}}}}}}$. Find the integral part of $S$.

Thanks in advance!

Answer 1

First, by numeric calculation, it is easy to find that S>2012, we only have to calculate some very small number of terms to get the result.

But for S<2013, it is hard to show that even with Mathematica, it will take a very long time. I cannot find a good method, but I checked that your conclusion is right.

$$log(S)=\Sigma_{i=0}^{2012^2-2012}\frac{1}{2^{i+1}}log(2012+i)$$. Then we can do the sum for the first few N=10000 terms with Mathematica,.Then for the other term with i larger than 10000, we can approximate the sum with integral:

$$\Sigma_{i=10000}^{2012^2-2012}\frac{1}{2^{i+1}}log(2012+i)\approx \int_{10000}^{2012^2-2012}dx \frac{1}{2^{x+1}}log(x+2012)$$ which is quite small. It's much smaller than $log(2013)$ minus the sum from 0 to N, thus your conclusion is proved to be true.

Answer 2

This is a partial answer!

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Define $$ S(n) := \sqrt{n \sqrt{(n+1) \dotsm \sqrt{n^2}}} = \prod_{k=0}^{n(n-1)} (n+k)^{2^{-k-1}}. $$ So far I can only prove that $$ n^2 \leq S(n)^2 $$ for every $n \geq 1$. Indeed, first observe that this is trivial for $n = 1$, so let's assume $n \geq 2$. Then note that $$ S(n)^2 = \prod_{k=0}^{n(n-1)} (n+k)^{2^{-k}} \geq n \prod_{k=1}^{n(n-1)} (n+1)^{2^{-k}} = n (n+1)^{e_n} $$ where $$ e_n = \sum_{k = 1}^{n(n-1)} \frac{1}{2^k} = 1 - 2^{-n^2 + n}. $$ Thus we need to prove that $$ \log(n) \leq e_n \log(n+1) $$ or, in other words $$ 2^{-n^2+n} \log(n+1) \leq \log(n+1) - \log(n) = \log\left(1 + \frac{1}{n}\right) $$ and rearranging the terms this becomes $$ 0 \leq \left(2^{n^2-n} - 1\right) \log\left(1 + \frac{1}{n}\right) - \log(n). $$ The derivative of the rhs is $$ 2^{n^{2} - n} {\left(2 \, n - 1\right)} \log\left(2\right) \log\left(\frac{1}{n} + 1\right) - \frac{2^{n^{2} - n} - 1}{n (n+1)} - \frac{1}{n} $$ which (squinting a bit) can be seen to be positive for every $n \geq 2$. Finally $$ \left(2^{2^2-2} - 1\right) \log\left(1 + \frac{1}{2}\right) - \log(2) = 3\log(3) - 4\log(2) = \log\left(\frac{27}{16}\right) > 0 $$ is enough to conclude that $n^2 \leq S(n)^2$ for every $n \geq 2$.

Answer 3

On the one hand, obviously $(2012^2-1)\sqrt{2012^2}>2012^2$, so we have \begin{align} \sqrt{(2012^2-1)\sqrt{2012^2}}&>2012\\ \sqrt{(2012^2-2)\sqrt{(2012^2-1)\sqrt{2012^2}}}&>2012, \\ \sqrt{(2012^2-3)\sqrt{(2012^2-2)\sqrt{(2012^2-1)\sqrt{2012^2}}}} >\sqrt{(2012^2-3)\cdot 2012}&>2012, \\ \sqrt{(2012^2-4)\sqrt{(2012^2-3)\sqrt{(2012^2-2)\sqrt{(2012^2-1)\sqrt{2012^2}}}}} >\sqrt{(2012^2-4)\cdot 2012}&>2012, \\ &\;\,\vdots \\ M>\sqrt{2012\cdot 2012}&=2012; \end{align} On the other hand, we have \begin{align} \sqrt{(2012^2-1)\sqrt{2012^2}}&<2012^2, \\ \sqrt{(2012^2-2)\sqrt{(2012^2-1)\sqrt{2012^2}}} <\sqrt{(2012^2-2)\cdot 2012^2}&<2012^2-1, \\ \sqrt{(2012^2-3)\sqrt{(2012^2-2)\sqrt{(2012^2-1)\sqrt{2012^2}}}} <\sqrt{(2012^2-3)(2012^2-1)}&<2012^2-2, \\ \sqrt{(2012^2-4)\sqrt{(2012^2-3)\sqrt{(2012^2-2)\sqrt{(2012^2-1)\sqrt{2012^2}}}}}<\sqrt{(2012^2-4)(2012^2-2)}&<2012^2-3, \\ &\;\;\vdots \\ M<\sqrt{2012\cdot 2014}&<2013. \end{align} In conclusion, the integer part of $M$ is $2012$.

(Reposted from Math&Recreation forum.)