The value of \(\lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{2n-1} \frac{n^2}{n^2 + 4r^2}\) is :
Show Hint
When the summation index \(r\) goes from \(0\) to \(kn-1\), the upper limit of the definite integral is \(k\). Always ensure you factor out \(1/n\) to correctly set up the Riemann sum.
Step 1: Understanding the Concept:
The limit of a sum as \(n \to \infty\) can be evaluated as a definite integral by expressing the sum in the form \(\frac{1}{n} \sum f\left(\frac{r}{n}\right)\). Here, the sum becomes an integral from the lower limit to the upper limit of the index variable. Step 2: Key Formula or Approach:
1. \(\lim_{n \to \infty} \frac{1}{n} \sum_{r=an}^{bn-1} f\left(\frac{r}{n}\right) = \int_a^b f(x) dx\).
2. \(\int \frac{1}{1+k^2x^2} dx = \frac{1}{k} \tan^{-1}(kx) + C\). Step 3: Detailed Explanation:
The given limit is:
\[ L = \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{2n-1} \frac{n^2}{n^2 + 4r^2} \]
Factor out \(n^2\) from the denominator:
\[ L = \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{2n-1} \frac{1}{1 + 4\left(\frac{r}{n}\right)^2} \]
This sum represents the definite integral:
\[ L = \int_0^2 \frac{1}{1 + 4x^2} dx \]
Integrating with respect to \(x\):
\[ L = \left[ \frac{1}{2} \tan^{-1}(2x) \right]_0^2 \]
Substituting the limits:
\[ L = \frac{1}{2} \left( \tan^{-1}(4) - \tan^{-1}(0) \right) = \frac{1}{2} \tan^{-1}(4) \] Step 4: Final Answer:
The value of the limit is \(\frac{1}{2} \tan^{-1}(4)\).
