Limits of a Composition

Observe that we can convert the limit into an exponential term that involves an integral as follows.

$\begin{aligned} A & = & \lim _{ n\to \infty } \left( 1+\frac { 1 }{ n^{ 2 } } \right) \left( \sqrt { 1+\frac { 4 }{ n^{ 2 } } } \right) \left( \sqrt [ 3 ]{ 1+\frac { 9 }{ n^{ 2 } } } \right) \cdots \left( \sqrt [ n ]{ 2 } \right)\\ &=& \lim _{ n\to \infty } \left( 1+\frac { 1 }{ n^{ 2 } } \right) \left( \sqrt { 1+\frac { 2^2 }{ n^{ 2 } } } \right) \left( \sqrt [ 3 ]{ 1+\frac { 3^2 }{ n^{ 2 } } } \right) \cdots \left( \sqrt [ n ]{ 1+\frac{n^2}{n^2} } \right)\\ &=&\exp\left(\lim _{ n\to \infty }\sum^{n}_{k=1}\frac{1}{k}\ln\left(1+\frac{k^2}{n^2}\right)\right)\\ &=&\exp\left(\lim _{ n\to \infty }\frac{1}{n}\sum^{n}_{k=1}\frac{n}{k}\ln\left(1+\frac{k^2}{n^2}\right)\right)\\ &=&\exp\left(\int^{1}_{0}\frac{\ln(1+x^2)}{x}\ dx\right) \end{aligned}$

Now, let us focus on the integral. Expanding the integrand about $x=0$ and evaluating gives

$\begin{aligned} \int^1_0\frac{\ln(1+x^2)}{x}\ dx&=&\int^1_0x-\frac{x^3}{2}+\frac{x^5}{3}-\frac{x^7}{4}+\cdots\ dx\\ &=&\left[\frac{x^2}{2}-\frac{x^4}{2\times4}+\frac{x^6}{3\times6}-\frac{x^8}{4\times8}+\cdots\right]^1_0\\ &=&\frac{1}{1\times2}-\frac{1}{2\times4}+\frac{1}{3\times6}-\frac{1}{4\times8}+\cdots\\ &=&\frac{1}{2}\left(\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\cdots\right)\\ &=&\frac{1}{2}\times\frac{\pi^2}{12}\ \text{(See note.)}\\ &=&\frac{\pi^2}{24}. \end{aligned}$

Putting everything together, we see that $A=e^{\frac{\pi^2}{24}}.$

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Note: Evaluation of the sum $S=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\cdots (*).$

We know that $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots=\zeta(2)=\frac{\pi^2}{6}.$ Then, we can manipulate the sum $(*)$ as follows. $\begin{aligned} S&=&\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\cdots\\ &=&{\color{#3D99F6}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots\right)}-{\color{magenta}2\left(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\cdots\right)}\\ &=&{\color{#3D99F6}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots\right)}-{\color{magenta}\frac{1}{2}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots\right)}\\ &=&\frac{1}{2}\left(\frac{\pi^2}{6}\right)\\ &=&\frac{\pi^2}{12}. \end{aligned}$

Thus, $S=\frac{\pi^2}{12}.$

It may not seem like it, but there is a reason why this problem was titled Limits of a Composition.

$A = \displaystyle \lim\limits_{n\to \infty} \bigg( 1 + \frac{1}{n^2}\bigg)\bigg( \sqrt{1 + \frac{4}{n^2}}\bigg)\bigg( \sqrt[3]{1 + \frac{9}{n^2}}\bigg)\bigg( \sqrt[4]{1 + \frac{16}{n^2}}\bigg)...\bigg(\sqrt[n]{2}\bigg)$

If we take the natural logarithm of both sides, we get the following:

$\ln A = \ln \Bigg[ \displaystyle \lim\limits_{n\to \infty} \bigg( 1 + \frac{1}{n^2}\bigg)\bigg( \sqrt{1 + \frac{4}{n^2}}\bigg)\bigg( \sqrt[3]{1 + \frac{9}{n^2}}\bigg)\bigg( \sqrt[4]{1 + \frac{16}{n^2}}\bigg)...\bigg(\sqrt[n]{2}\bigg) \Bigg]$

Now, for any two functions $f(x)$ and $g(x)$ such that $\lim\limits_{x \to a} g(x) = b$ , and that $f$ is continuous at $b$ , then we have the following relationship:

$\lim\limits_{x \to a} f ( g(x) ) = f( \lim\limits_{x \to a} g(x))$

For our problem, the long equation equating to $A$ is $g(x)$ , while $f(x) = \ln x$ . This will now enable us to remove the limit operator inside the logarithm operand.

It now becomes..

$\ln A = \displaystyle \lim\limits_{n\to \infty} \ln \Bigg[ \bigg( 1 + \frac{1}{n^2}\bigg)\bigg( \sqrt{1 + \frac{4}{n^2}}\bigg)\bigg( \sqrt[3]{1 + \frac{9}{n^2}}\bigg)\bigg( \sqrt[4]{1 + \frac{16}{n^2}}\bigg)...\bigg(\sqrt[n]{2}\bigg) \Bigg]$

Hence the title.

$\ln A = \displaystyle \lim\limits_{n \to \infty} \Big[ \ln(1 + \frac{1}{n^2})+ \frac{ \ln(1 + \frac{4}{n^2})}{2} + \frac{\ln(1 + \frac{9}{n^2})}{3}+ ... \frac{\ln 2}{n} \Big]$

$\ln A = \displaystyle \lim\limits_{n \to \infty} \sum\limits_{k=1}^{\infty} \frac{ \ln (1 + \frac{k^2}{n^2})}{k}$

A little bit of modification yields

$\ln A = \displaystyle \lim\limits_{n \to \infty} \frac{1}{n} \sum\limits_{k=1}^{\infty} \frac{ \ln (1 + \frac{k^2}{n^2})}{\frac{k}{n}}$

$\ln A = \displaystyle \int_0^1 \frac{ \ln(1 + x^2)}{x}$

By a similar method used to solve this problem , we find out that $\ln A = \frac{\pi^2}{24}$ .

And so we find $\displaystyle A = e^{\frac{\pi^2}{24}}$ .

And... $\lfloor 10^4 A \rfloor = \boxed { 15086 }$ .