Given that `lim_(nrarroo) sum_(r=1)^(n) (log_(e)(n^(2)+r^(2))-2log_(e)n)/n = log_(e)2+pi/2-2`, then
evaluate : ` lim_(nrarroo) (1)/(n^(2m))[(n^(2)+1^(2))^(m)(n^(2)+2^(2))^(m) "......"(2n^(2))^(m)]^(1//n)`.

Answer

To solve the given limit problem, we will break it down step by step.

**Given:**
\[ \lim_{n \to \infty} \sum_{r=1}^{n} \left( \log(n^2 + r^2) - 2 \log n \right) / n = \log 2 + \frac{\pi}{2} - 2 \]

We need to evaluate:
\[ \lim_{n \to \infty} \frac{1}{n^{2m}} \left[ (n^2 + 1^2)^m (n^2 + 2^2)^m \cdots (n^2 + (2n)^2)^m \right]^{\frac{1}{n}} \]

**Step 1: Simplifying the expression inside the limit**
We can rewrite the expression inside the limit:
\[ \frac{1}{n^{2m}} \left[ (n^2 + 1^2)(n^2 + 2^2) \cdots (n^2 + (2n)^2) \right]^{\frac{m}{n}} \]

**Step 2: Analyzing the product**
The product can be expressed as:
\[ (n^2 + 1^2)(n^2 + 2^2) \cdots (n^2 + (2n)^2) = \prod_{r=1}^{2n} (n^2 + r^2) \]

**Step 3: Taking the logarithm**
To simplify the limit, we can take the logarithm of the product:
\[ \log \left( \prod_{r=1}^{2n} (n^2 + r^2) \right) = \sum_{r=1}^{2n} \log(n^2 + r^2) \]

**Step 4: Approximating the logarithm**
As \( n \to \infty \), we can approximate:
\[ \log(n^2 + r^2) \approx \log(n^2) + \log\left(1 + \frac{r^2}{n^2}\right) \approx 2\log n + \frac{r^2}{n^2} - \frac{r^4}{2n^4} + O\left(\frac{1}{n^6}\right) \]

**Step 5: Summing the logarithms**
Now, we can sum this approximation:
\[ \sum_{r=1}^{2n} \log(n^2 + r^2) \approx \sum_{r=1}^{2n} \left( 2 \log n + \frac{r^2}{n^2} \right) \]

The first part gives:
\[ 2n \log n \]
The second part can be approximated using the formula for the sum of squares:
\[ \sum_{r=1}^{2n} r^2 = \frac{(2n)(2n + 1)(4n + 1)}{6} \approx \frac{4n^3}{6} = \frac{2n^3}{3} \]
Thus,
\[ \sum_{r=1}^{2n} \frac{r^2}{n^2} \approx \frac{2n^3/3}{n^2} = \frac{2n}{3} \]

**Step 6: Putting it all together**
Combining these results:
\[ \sum_{r=1}^{2n} \log(n^2 + r^2) \approx 2n \log n + \frac{2n}{3} \]

**Step 7: Final limit calculation**
Now we can substitute back into our limit:
\[ \lim_{n \to \infty} \frac{1}{n^{2m}} \left[ e^{2n \log n + \frac{2n}{3}} \right]^{\frac{m}{n}} = \lim_{n \to \infty} \frac{1}{n^{2m}} e^{2m \log n + \frac{2m}{3}} = \lim_{n \to \infty} \frac{e^{\frac{2m}{3}}}{n^{2m - 2m}} = e^{\frac{2m}{3}} \]

Thus, we conclude:
\[ \lim_{n \to \infty} \frac{1}{n^{2m}} \left[ (n^2 + 1^2)^m (n^2 + 2^2)^m \cdots (n^2 + (2n)^2)^m \right]^{\frac{1}{n}} = e^{\frac{2m}{3}} \]

**Final Answer:**
\[ e^{\frac{2m}{3}} \]

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