Definite Integration
Grade None
Question:
<p>If L = <span class="math-tex">\(\lim _\limits{n \rightarrow \infty} \frac{1}{n^{4}} \sum_\limits{i=1}^{2 n}\left(n^{2}+i^{2}\right)^{\frac{1}{n}}\)</span>, then ln L is equal to :</p>
<p style="display:inline">2 tan<sup>-1</sup> 2 + 4 + 2 ln 5</p>
<p style="display:inline">2 tan<sup>-1</sup> 2 - 4 + 2 ln 5</p>
<p style="display:inline">2 ln 5 - 4 - 2 tan<sup>-1</sup> 2</p>
<p style="display:inline">ln <span class="math-tex">\(2+\frac{\pi}{2}-2\)</span></p>
Step-by-Step Solution
Key Concept: Convert the limit of the expression into a definite integral by using properties of logarithms to transform the product into a Riemann sum of the form $\frac{1}{n} \sum f(i/n)$.
<p>We have <span class="math-tex">\(L=\lim \limits_{n \rightarrow \infty} \frac{1}{n^{4}} \sum \limits_{i=1}^{2 n}\left(n^{2}+i^{2}\right)^{1 / n}\)</span><br />
<span class="math-tex">\(\Rightarrow \ln L=\lim \limits_{n \rightarrow \infty} \frac{1}{n} \sum \limits_{i=1}^{2 n} \ln \left(n^{2}+i^{2}\right)-4 \ln n\)</span><br />
<span class="math-tex">\(\Rightarrow \ln L=\lim \limits_{n \rightarrow \infty} \frac{1}{n} \sum \limits_{i=1}^{2 n} \ln \left(n^{2}\left(1+\frac{i^{2}}{n^{2}}\right)\right)-4 \ln n\)</span><br />
<span class="math-tex">\(\Rightarrow \ln L=\lim \limits_{n \rightarrow \infty} \frac{2 \ln n}{n} \sum \limits_{i=1}^{2 n} 1+\frac{1}{n} \sum \limits_{i=1}^{2 n} \ln \left(1+\frac{i^{2}}{n^{2}}\right)-4 \ln n\)</span><br />
<span class="math-tex">\(\Rightarrow \ln L=\lim \limits_{n \rightarrow \infty} \frac{2 \ln n}{n}-(2 n)+\frac{1}{n} \sum \limits_{i=1}^{2 n} \ln \left(1+\frac{i^{2}}{n^{2}}\right)-4 \ln n\)</span><br />
<span class="math-tex">\(\Rightarrow \ln L=\int_{0}^{2} \ln \left(1+x^{2}\right) d x\)</span><span class="math-tex">\(=\left.x \ln \left(1+x^{2}\right)\right|_{0} ^{2}-\int_{0}^{2} \frac{2 x^{2}}{1+x^{2}} d x\)</span><br />
So, In L = 2 In 5 - 2[2 - tan<sup>-1</sup> 2] = 2tan<sup>-1</sup> 2 - 4 + 2 In 5</p>
Correct Answer: B