Key Concepts and Formulas

• Limits of Products and Logarithms: When evaluating limits of expressions involving products, taking the natural logarithm is a standard technique to convert the product into a sum. This is because $\log(\prod a_i) = \sum \log(a_i)$ and $\log(a^b) = b \log a$.
• Riemann Sums and Definite Integrals: A limit of a sum of the form $\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 1}^n {f\left( {{r \over n}} \right)}$ can be converted into a definite integral $\int_0^1 f(x) \, dx$. Here, $x$ corresponds to $\frac{r}{n}$ and $dx$ corresponds to $\frac{1}{n}$.
• Integration by Substitution: This technique is used to simplify integrals by substituting a part of the integrand with a new variable. If $u = g(x)$, then $du = g'(x) \, dx$. The limits of integration also need to be changed according to the substitution.

Step-by-Step Solution

Step 1: Express $U_n$ and the Limit in a Usable Form

We are given ${U_n} = \left( {1 + {1 \over {{n^2}}}} \right)\left( {1 + {{{2^2}} \over {{n^2}}}} \right)^2.....\left( {1 + {{{n^2}} \over {{n^2}}}} \right)^n$. This can be written using product notation as: ${U_n} = \prod\limits_{r = 1}^n {{{\left( {1 + {{{r^2}} \over {{n^2}}}} \right)}^r}}$ We need to find the limit $L = \mathop {\lim }\limits_{n \to \infty } {({U_n})^{{{ - 4} \over {{n^2}}}}}$

Step 2: Apply Natural Logarithm to Simplify the Limit

To handle the limit of an expression raised to a power, we take the natural logarithm of the entire expression. $\log L = \mathop {\lim }\limits_{n \to \infty } \log \left( {({U_n})^{{{ - 4} \over {{n^2}}}}} \right)$ Using the logarithm property $\log(a^b) = b \log a$: $\log L = \mathop {\lim }\limits_{n \to \infty } {{ - 4} \over {{n^2}}} \log ({U_n})$ Substitute the product form of $U_n$: $\log L = \mathop {\lim }\limits_{n \to \infty } {{ - 4} \over {{n^2}}} \log \left( {\prod\limits_{r = 1}^n {{{\left( {1 + {{{r^2}} \over {{n^2}}}} \right)}^r}} } \right)$ Using the logarithm property $\log(\prod a_i) = \sum \log(a_i)$: $\log L = \mathop {\lim }\limits_{n \to \infty } {{ - 4} \over {{n^2}}} \sum\limits_{r = 1}^n {\log {{\left( {1 + {{{r^2}} \over {{n^2}}}} \right)}^r}}$ Applying $\log(a^b) = b \log a$ again: $\log L = \mathop {\lim }\limits_{n \to \infty } {{ - 4} \over {{n^2}}} \sum\limits_{r = 1}^n {r \log \left( {1 + {{{r^2}} \over {{n^2}}}} \right)}$

Step 3: Rearrange the Sum for Riemann Sum Formation

We need to convert the sum into the form $\frac{1}{n}\sum f(\frac{r}{n})$. Let's rearrange the terms: $\log L = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {\left( {{{ - 4r} \over {{n^2}}}} \right) \log \left( {1 + {{{r^2}} \over {{n^2}}}} \right)}$ To achieve the $\frac{1}{n}$ factor, we write $\frac{r}{n^2}$ as $\frac{r}{n} \cdot \frac{1}{n}$: $\log L = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n { - 4 \left( {{r \over n}} \right) \left( {{1 \over n}} \right) \log \left( {1 + {{\left( {{r \over n}} \right)}^2}} \right)}$

Step 4: Convert the Sum to a Definite Integral

Now, we can identify the components of a Riemann sum:

• $x = \frac{r}{n}$
• $dx = \frac{1}{n}$
• The function is $f(x) = -4x \log(1+x^2)$.
• As $n \to \infty$, $r$ goes from $1$ to $n$, so $\frac{r}{n}$ goes from $\frac{1}{n} \to 0$ to $\frac{n}{n} = 1$. The limits of integration are from $0$ to $1$.

Thus, the limit of the sum becomes a definite integral: $\log L = \int\limits_0^1 { - 4x\log (1 + {x^2})\,dx}$ $\log L = - 4\int\limits_0^1 {x\log (1 + {x^2})\,dx}$

Step 5: Evaluate the Definite Integral

We use integration by substitution. Let $u = 1 + x^2$. Then, $du = 2x \, dx$, which means $x \, dx = \frac{1}{2} du$.

We also need to change the limits of integration: When $x = 0$, $u = 1 + 0^2 = 1$. When $x = 1$, $u = 1 + 1^2 = 2$.

Substituting these into the integral: $\log L = - 4\int\limits_1^2 {\log(u) \left( {{1 \over 2}}\,du \right)}$ $\log L = - 4 \cdot {1 \over 2} \int\limits_1^2 {\log(u)\,du}$ $\log L = - 2 \int\limits_1^2 {\log(u)\,du}$

The integral of $\log(u)$ is $u \log(u) - u$. $\log L = - 2 \left[ {u \log(u) - u} \right]_1^2$ Now, we evaluate at the limits: $\log L = - 2 \left[ {(2 \log(2) - 2) - (1 \log(1) - 1)} \right]$ Since $\log(1) = 0$: $\log L = - 2 \left[ {2 \log(2) - 2 - (0 - 1)} \right]$ $\log L = - 2 \left[ {2 \log(2) - 2 + 1} \right]$ $\log L = - 2 \left[ {2 \log(2) - 1} \right]$ $\log L = - 4 \log(2) + 2$

Step 6: Solve for $L$

We have $\log L = 2 - 4 \log(2)$. Using logarithm properties, $4 \log(2) = \log(2^4) = \log(16)$. So, $\log L = 2 - \log(16)$. We can write $2$ as $\log(e^2)$. $\log L = \log(e^2) - \log(16)$ Using the property $\log a - \log b = \log(\frac{a}{b})$: $\log L = \log\left( {{e^2} \over {16}} \right)$ Therefore, $L = {{e^2} \over {16}}$.

Common Mistakes & Tips

• Incorrect Riemann Sum Setup: Ensure that the expression inside the sum can be correctly transformed into the form $f(\frac{r}{n}) \cdot \frac{1}{n}$. Pay close attention to how powers of $n$ are distributed.
• Logarithm Properties: Mistakes in applying $\log(a^b) = b \log a$ or $\log(\prod a_i) = \sum \log(a_i)$ can lead to incorrect sums.
• Integration by Substitution Errors: Incorrectly changing the limits of integration or the differential element ($du$) when performing substitution is a common pitfall.

Summary

The problem requires converting a limit of a power of a product into a definite integral. This is achieved by taking the natural logarithm to transform the product into a sum and then recognizing the sum as a Riemann sum. The resulting definite integral is evaluated using integration by substitution. The final step involves exponentiating the result of the logarithm to find the value of the original limit.

The final answer is \boxed{\frac{e^2}{16}}. This corresponds to option (A).