Whenever you see a limit of the form $\frac{1}{n}\sum_{k=1}^n a_k$, immediately check $\lim_{n \to \infty} a_n$. If $a_n \to L$, then the average also converges to $L$ by Cauchy's First Limit Theorem!
Step 1: Concept
This problem involves calculating the limit of the arithmetic mean of a sequence $a_n = \frac{1}{n}$. We can apply Cauchy's First Theorem on Limits. Step 2: Key Formulas and Approach Cauchy's First Theorem on Limits: If a sequence $\langle a_n \rangle$ converges to a limit $L$, i.e., $\lim_{n \to \infty} a_n = L$, then the sequence of arithmetic means $\langle x_n \rangle$ defined by:
\[ x_n = \frac{a_1 + a_2 + a_3 + \dots + a_n}{n} \]
also converges to $L$, i.e., $\lim_{n \to \infty} x_n = L$. Step 3: Step-by-step Explanation
• Define $a_n = \frac{1}{n}$.
• Evaluate the limit of $a_n$ as $n \to \infty$:
\[ L = \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{n} = 0 \]
• Now construct the sequence of arithmetic means $x_n$:
\[ x_n = \frac{a_1 + a_2 + a_3 + \dots + a_n}{n} = \frac{1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}}{n} \]
• By Cauchy's First Theorem on Limits, since $\lim_{n \to \infty} a_n = 0$, we immediately have:
\[ \lim_{n \to \infty} x_n = \lim_{n \to \infty} \frac{1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}}{n} = 0 \]
• Alternative Method (Using Asymptotic Bounds):
It is well known that $H_n = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} \approx \ln(n) + \gamma$, where $\gamma$ is Euler-Mascheroni constant.
Therefore:
\[ \lim_{n \to \infty} \frac{H_n}{n} = \lim_{n \to \infty} \frac{\ln(n) + \gamma}{n} = 0 \]
both methods yield the exact same result $0$. Step 4: Final Answer
The limit equals 0. Thus, Option (B) is correct.
