Question

The value of the limit \( \lim_{n \to \infty} \left( \frac{1}{1^2 + n^2} + \frac{2}{2^2 + n^2} + \frac{3}{3^2 + n^2} + \cdots + \frac{n}{n^2 + n^2} \right) \) is equal to:

• A. 1
• B. \( \frac{1}{2}\log 2 \)
• C. \( 2\log 2 \)
• D. 0

Hint:

For limits of sums, always look for the \( \frac{1}{n} f(\frac{r}{n}) \) pattern. If the power of \( n \) in the denominator is one higher than the power of \( r \) in the numerator across the whole expression, it's a prime candidate for a Riemann sum integral.

Answer 1

The Correct Option is B

Concept: A limit of a sum can often be converted into a definite integral using the definition of the Riemann sum. The general transformation is:
• Identify the general term \( T_r \).
• Express the sum in sigma notation: \( \lim_{n \to \infty} \sum_{r=1}^{n} T_r \).
• Replace \( \frac{r}{n} \) with \( x \), \( \frac{1}{n} \) with \( dx \), and the summation with \( \int_{0}^{1} \).

Step 1: Writing the sum in sigma notation.
The general term of the series is \( \frac{r}{r^2 + n^2} \). We can write the limit as: \[ L = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{r}{r^2 + n^2} \] To prepare for the integral transformation, we divide the numerator and denominator by \( n^2 \): \[ L = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{r/n^2}{(r^2/n^2) + 1} = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n} \cdot \frac{r/n}{(r/n)^2 + 1} \]

Step 2: Converting the limit of the sum into a definite integral.
Using the transformations \( \frac{r}{n} \to x \) and \( \frac{1}{n} \to dx \): \[ L = \int_{0}^{1} \frac{x}{x^2 + 1} \, dx \]

Step 3: Evaluating the definite integral.
To integrate \( \frac{x}{x^2 + 1} \), we use the substitution \( u = x^2 + 1 \), which gives \( du = 2x \, dx \Rightarrow x \, dx = \frac{du}{2} \). \[ L = \frac{1}{2} \int \frac{du}{u} = \frac{1}{2} [ \log|x^2 + 1| ]_{0}^{1} \] Substituting the limits: \[ L = \frac{1}{2} [ \log(1^2 + 1) - \log(0^2 + 1) ] \] \[ L = \frac{1}{2} [ \log 2 - \log 1 ] = \frac{1}{2} \log 2 \quad (\because \log 1 = 0) \]