Question

\[ \lim_{n\to\infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{4}{n^2}\right)\left(1+\frac{9}{n^2}\right)\cdots\left(1+\frac{n^2}{n^2}\right)\right]^{\frac{1}{n}} = \]

• A. \(16e^{-1}\)
• B. \(2e^{\left(\frac{\pi-4}{2}\right)}\)
• C. \(2\log 2 - 1\)
• D. \(2+e^{\left(\frac{\pi-4}{2}\right)}\)

Hint:

When evaluating limits of products, taking the logarithm can turn the product into a sum, which can often be expressed as a Riemann sum.

Answer 1

The Correct Option is B

Let \( L \) be the limit. Then \[ \ln L = \lim_{n \to \infty} \frac{1}{n} \ln \left[ \prod_{k=1}^n \left( 1 + \frac{k^2}{n^2} \right) \right] \] \[ = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \ln \left( 1 + \frac{k^2}{n^2} \right) \] \[ = \int_0^1 \ln (1 + x^2) \, dx. \] We integrate by parts, with \( u = \ln (1 + x^2) \) and \( dv = dx \), so \( du = \frac{2x}{1 + x^2} \, dx \) and \( v = x \). Then \[ \int_0^1 \ln (1 + x^2) \, dx = x \ln (1 + x^2) \bigg|_0^1 - \int_0^1 \frac{2x^2}{1 + x^2} \, dx \] \[ = \ln 2 - 2 \int_0^1 \frac{x^2}{1 + x^2} \, dx \] \[ = \ln 2 - 2 \int_0^1 \frac{1 + x^2 - 1}{1 + x^2} \, dx \] \[ = \ln 2 - 2 \int_0^1 \left( 1 - \frac{1}{1 + x^2} \right) \, dx \] \[ = \ln 2 - 2 (x - \arctan x) \bigg|_0^1 \] \[ = \ln 2 - 2 \left( 1 - \frac{\pi}{4} \right) \] \[ = \ln 2 - 2 + \frac{\pi}{2}. \] Then \[ L = e^{\ln 2 - 2 + \frac{\pi}{2}} = e^{\ln 2} e^{-2 + \frac{\pi}{2}} = 2 e^{\frac{\pi - 4}{2}}. \]