Question

\(\lim_{n \to \infty} \frac{1}{n^2} \left[ e^{1/n} + 2e^{2/n} + 3e^{3/n} + \dots + 2n e^{2n/n} \right] =\)

• A. \(e^2 - 1\)
• B. \(e^2 + 1\)
• C. \(2e^2 - 2\)
• D. \(2e^2 + 1\)

Hint:

When converting a summation to an integral \(\lim_{n \to \infty} \frac{1}{n} \sum f(r/n)\), always check the upper limit of summation. If it goes up to \(kn\), the integral upper bound is \(k\).

Answer 1

The Correct Option is B

Step 1: Express as a Riemann Sum:
The given limit is: \[ L = \lim_{n \to \infty} \sum_{r=1}^{2n} \frac{r}{n^2} e^{r/n} \] Rewrite \(\frac{r}{n^2}\) as \(\frac{1}{n} \cdot \frac{r}{n}\): \[ L = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{2n} \left( \frac{r}{n} \right) e^{r/n} \]
Step 2: Convert to Definite Integral:
Let \(x = \frac{r}{n}\) and \(dx = \frac{1}{n}\). Lower limit: As \(r \to 1, n \to \infty, x \to 0\). Upper limit: As \(r \to 2n, n \to \infty, x \to 2\). \[ L = \int_{0}^{2} x e^x dx \]
Step 3: Solve the Integral:
Using Integration by Parts (\(\int u dv = uv - \int v du\)): Let \(u = x \implies du = dx\). Let \(dv = e^x dx \implies v = e^x\). \[ \int_{0}^{2} x e^x dx = \left[ x e^x \right]_0^2 - \int_{0}^{2} e^x dx \] \[ = (2e^2 - 0e^0) - [e^x]_0^2 \] \[ = 2e^2 - (e^2 - e^0) \] \[ = 2e^2 - (e^2 - 1) \] \[ = e^2 + 1 \]