Definite Integration
Limit of Sum — Advanced
Grade 12
Question:
<p>If \(\displaystyle\lim_{n\to\infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{na}\right)=1\), find \(a\). [JEE Advanced 2013]</p>
\(e\)
\(1/e\)
\(e^2\)
\(2e\)
Step-by-Step Solution
Key Concept: Riemann sum \to \int_1^a dx/x = ln a. Set ln a = 1 \to a = e.
<div class='solution'>
<p>$$\sum_{k=1}^{n(a-1)}\frac{1}{n+k}=\frac{1}{n}\sum_{k=1}^{n(a-1)}\frac{1}{1+k/n}\to\int_0^{a-1}\frac{dx}{1+x}=\ln a$$</p>
<p>Setting $\ln a=1\Rightarrow a=\boxed{e}$.</p>
Correct Answer: A