Question

Evaluate the limit: \[ \lim_{n \to \infty} \left( \frac{1}{1^2 + n^2} + \frac{2}{2^2 + n^2} + \frac{3}{3^2 + n^2} + \cdots + \frac{n}{n^2 + n^2} \right) \]

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

Hint:

When you see sums with patterns like \(\frac{r}{r^2 + n^2}\), think about rewriting them as Riemann sums to evaluate limits using definite integrals.

Answer 1

The Correct Option is B

We observe the expression inside the limit is a Riemann sum: \[ \sum_{r=1}^{n} \frac{r}{r^2 + n^2} \] Divide numerator and denominator by \(n^2\): \[ = \sum_{r=1}^{n} \frac{r/n}{(r/n)^2 + 1} \cdot \frac{1}{n} \] This is a Riemann sum approximation for the integral: \[ \int_0^1 \frac{x}{x^2 + 1} dx \] To solve: Let \( I = \int_0^1 \frac{x}{x^2 + 1} dx \) Use substitution: Let \( u = x^2 + 1 \Rightarrow du = 2x dx \) Then: \[ I = \frac{1}{2} \int_{x=0}^{1} \frac{2x}{x^2 + 1} dx = \frac{1}{2} \int_{u=1}^{2} \frac{1}{u} du = \frac{1}{2} \ln u \Big|_{1}^{2} = \frac{1}{2} \log 2 \] Hence, the limit evaluates to: \[ \boxed{\dfrac{1}{2} \log 2} \]