If $\sum_{r=1}^{n} a_r = \sum_{k=1}^{n} \sum_{j=1}^{k} 2$ and $\lambda = \lim_{n \to \infty} \left(\sum_{i=1}^{n} \frac{1}{a_i}\right)$, then $\left[\frac{1}{\lambda}\right]$ is equal to (where $[.]$ denotes greatest integer function).
Step-by-Step Solution
Key Concept: Telescoping series simplifies complex sums by cancellation, and limit of $(1-1/n)^n$ equals $e^{-1}$.
The sum $S_n = \frac{1}{3}n(n+1)(n+2)$ is derived by expanding the telescoping series. The general term is $a_n = n(n+1)$, which can be verified by computing $S_n - S_{n-1}$. The sum of reciprocals $\sum_{r=1}^{n} \frac{1}{a_r} = \sum_{r=1}^{n} \left(\frac{1}{r} - \frac{1}{r+1}\right) = 1 - \frac{1}{n+1} = \frac{n}{n+1}$. As $n \to \infty$, this approaches 1, and using the limit $\lim_{n \to \infty} \left(\frac{n}{n+1}\right)^n = e^{-1}$, we find $\frac{1}{k} = e$, giving $\left[\frac{1}{k}\right] = 2$.
Correct Answer: 2