Question

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

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

Hint:

Recognize the given limit as a Riemann sum. The form \( \lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^n f\left(\frac{i}{n}\right) \) corresponds to the definite integral \( \int_0^1 f(x) dx \). In this case, \( f(x) = e^{-x} \). Evaluate this definite integral to find the value of the limit.

Answer 1

The Correct Option is A

The given limit is of the form of a Riemann sum.
$$ \lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^n \frac{1}{e^{i/n}} = \lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^n e^{-i/n} $$ This can be interpreted as the definite integral of the function \( f(x) = e^{-x} \) over the interval \( [0, 1] \).
The Riemann sum is given by \( \lim_{n \to \infty} \frac{b - a}{n} \sum_{i=1}^n f(a + i \frac{b - a}{n}) \).
Comparing this with the given limit, we have \( a = 0 \), \( b - a = 1 \), so \( b = 1 \), and \( f(x) = e^{-x} \).
Thus, the limit is equal to the definite integral: $$ \int_0^1 e^{-x} dx $$ Evaluating the integral: $$ \int_0^1 e^{-x} dx = [-e^{-x}]_0^1 = -e^{-1} - (-e^{-0}) = -e^{-1} - (-1) = 1 - e^{-1} $$