Sequences & Series
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Grade 11
Question:
Let $S_n$ where $n \in N$ be the sum of infinite geometric progression whose first term is $n$ and common ratio is $\frac{1}{n+1}$. If $\lim_{n\to\infty} \frac{S_1S_n + S_2S_{n-1} + S_3S_{n-2} + \dots + S_nS_1}{S_1^2 + S_2^2 + S_3^2 + \dots + S_n^2} = \frac{p}{q}$ where $p$ and
Step-by-Step Solution
Key Concept: sum of an infinite geometric progression and limit of a series
$$\begin{aligned}
\color{blue}{\text{Step 1:}} \quad L &= \lim_{n\to\infty} \frac{\sum_{k=1}^{n} S_k S_{n-k+1}}{\sum_{k=1}^{n} S_k^2} \\
\color{blue}{\text{Step 2:}} \quad \text{Now, } S_n &= \frac{n}{1 - \frac{1}{n+1}} = \frac{n(n+1)}{n} = n+1 \\
\color{blue}{\text{Step 3:}} \quad L &= \lim_{n\to\infty} \frac{\sum_{k=1}^{n} (k+1)(n-k+2)}{\sum_{k=1}^{n} (k+1)^2} \\
&= \lim_{n\to\infty} \frac{\sum_{k=1}^{n} (k+1)(n-(k+1)+3)}{\sum_{k=1}^{n} (k+1)^2} \\
&= \frac{1}{2} \\
\end{aligned}$$
Therefore: $\frac{1}{2}$
Correct Answer: 3