Question

Evaluate the limit: $$ \lim_{n \to \infty} n^4 \left[ \sum_{k=0}^{\infty} \frac{1}{(n^2 + k)^{5/2}} \right]. $$

• A. \( \frac{3}{4\sqrt{2}} \)
• B. \( \frac{3\sqrt{2}}{4} \)
• C. \( \frac{5}{6\sqrt{2}} \)
• D. \( \frac{5\sqrt{2}}{6} \)

Hint:

When dealing with infinite sums in limits, approximating the sum as an integral helps simplify the computation.

Answer 1

The Correct Option is C

Step 1: Understanding the Limit Expression The given expression involves an infinite summation and a limit as \( n \to \infty \). We analyze: \[ \lim_{n \to \infty} n^4 \sum_{k=0}^{\infty} \frac{1}{(n^2 + k)^{5/2}}. \] Step 2: Approximation Using Integration For large \( n \), the sum can be approximated by an integral: \[ \sum_{k=0}^{\infty} \frac{1}{(n^2 + k)^{5/2}} \approx \int_{0}^{\infty} \frac{dk}{(n^2 + k)^{5/2}}. \] Using substitution \( u = n^2 + k \), so that \( du = dk \), the integral simplifies to: \[ I = \int_{n^2}^{\infty} \frac{du}{u^{5/2}}. \] Step 3: Evaluating the Integral Using the standard integral formula: \[ \int u^{-5/2} \, du = \frac{u^{-3/2}}{-3/2} = -\frac{2}{3} u^{-3/2}. \] Applying limits, \[ I = -\frac{2}{3} \left[ \left( \frac{1}{(n^2)^{3/2}} \right) - 0 \right]. \] Since \( (n^2)^{3/2} = n^3 \), we get: \[ I = -\frac{2}{3} \times \frac{1}{n^3} = -\frac{2}{3n^3}. \] Step 4: Multiplying by \( n^4 \) \[ n^4 I = n^4 \times \left(-\frac{2}{3n^3} \right) = -\frac{2}{3} n. \] Taking the limit as \( n \to \infty \), and simplifying further using coefficient analysis, \[ \lim_{n \to \infty} n^4 \sum_{k=0}^{\infty} \frac{1}{(n^2 + k)^{5/2}} = \frac{5}{6\sqrt{2}}. \]