Question

If \(\lim_{n\rightarrow\infty}[(1+\frac{1}{n^{2}})(1+\frac{4}{n^{2}})(1+\frac{9}{n^{2}})\cdots(1+\frac{n^{2}}{n^{2}})]^{\frac{1}{n}}=ae^{b}\), then a+b=

• A. \(\pi-2\)
• B. \(\pi\)
• C. \(\pi+2\)
• D. \(\frac{\pi}{2}\)

Hint:

Recognize the limit as a Riemann sum and evaluate the integral using integration by parts.

Answer 1

The Correct Option is D

Step 1: Rewrite the given expression.
Let L = \(\lim_{n\rightarrow\infty}[(1+\frac{1}{n^{2}})(1+\frac{4}{n^{2}})(1+\frac{9}{n^{2}})\cdots(1+\frac{n^{2}}{n^{2}})]^{\frac{1}{n}}\).
L = \(\lim_{n\rightarrow\infty}\left[\prod_{k=1}^{n}\left(1+\frac{k^2}{n^2}\right)\right]^{\frac{1}{n}}\)

Step 2: Take logarithm on both sides.
\(\ln L = \lim_{n\rightarrow\infty} \frac{1}{n} \sum_{k=1}^{n} \ln\left(1+\frac{k^2}{n^2}\right)\)

Step 3: Recognize the Riemann sum.
\(\ln L = \lim_{n\rightarrow\infty} \frac{1}{n} \sum_{k=1}^{n} \ln\left(1+\left(\frac{k}{n}\right)^2\right)\)
This is a Riemann sum, so we can write it as an integral:
\(\ln L = \int_{0}^{1} \ln(1+x^2) \, dx\)

Step 4: Evaluate the integral using integration by parts.
Let u = \(\ln(1+x^2)\) and dv = dx.
Then du = \(\frac{2x}{1+x^2} \, dx\) and v = x.
\(\ln L = x \ln(1+x^2) \bigg|_{0}^{1} - \int_{0}^{1} \frac{2x^2}{1+x^2} \, dx\)
\(\ln L = \ln(2) - 2 \int_{0}^{1} \frac{x^2}{1+x^2} \, dx\)
\(\ln L = \ln(2) - 2 \int_{0}^{1} \frac{1+x^2 - 1}{1+x^2} \, dx\)
\(\ln L = \ln(2) - 2 \int_{0}^{1} \left(1 - \frac{1}{1+x^2}\right) \, dx\)
\(\ln L = \ln(2) - 2 \left[x - \arctan(x)\right]_{0}^{1}\)
\(\ln L = \ln(2) - 2 \left[(1 - \arctan(1)) - (0 - \arctan(0))\right]\)
\(\ln L = \ln(2) - 2 \left(1 - \frac{\pi}{4}\right)\)
\(\ln L = \ln(2) - 2 + \frac{\pi}{2}\)

Step 5: Compare with ae^b.
L = \(e^{\ln(2) - 2 + \frac{\pi}{2}} = e^{\ln(2)} e^{-2} e^{\frac{\pi}{2}} = 2 e^{\frac{\pi}{2} - 2}\)
So a = 2 and b = \(\frac{\pi}{2} - 2\).
a + b = \(2 + \frac{\pi}{2} - 2 = \frac{\pi}{2}\)
Therefore, a+b = \(\frac{\pi}{2}\).