Limits, Continuity & Differentiability
Limits of Series
Grade 12
Question:
<p>The value of \(\lim_{n \to \infty} \sum_{i=1}^{n} a_i\) is equal to</p>
<p>(a) \(\frac{1}{4}\)</p>
<p>(b) \(\frac{1}{2}\)</p>
<p>(c) \(1\)</p>
<p>(d) \(2\)</p>
Step-by-Step Solution
Key Concept: Use the method of multiplying by the common ratio and subtracting to find the sum of an arithmetico-geometric series, then take the limit as n approaches infinity.
<p><strong>Solution:</strong> Given that $S_n = \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \ldots + \frac{n}{2^n}$</p><p>Let $S_n = \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \ldots + \frac{n}{2^n}$ ... (i)</p><p>Then $\frac{S_n}{2} = \frac{1}{2^2} + \frac{2}{2^3} + \frac{3}{2^4} + \ldots + \frac{n}{2^{n+1}}$ ... (ii)</p><p>Subtracting (ii) from (i):</p><p>$S_n - \frac{S_n}{2} = \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \ldots + \frac{1}{2^n} - \frac{n}{2^{n+1}}$</p><p>$\frac{S_n}{2} = \frac{1/2(1-(1/2)^n)}{1-1/2} - \frac{n}{2^{n+1}} = 1 - \frac{1}{2^n} - \frac{n}{2^{n+1}}$</p><p>$S_n = 2\left(1 - \frac{1}{2^n} - \frac{n}{2^{n+1}}\right)$</p><p>As $n \to \infty$, $\lim_{n \to \infty} S_n = 2$</p><p>∴ Answer is (d) 2.</p>
Correct Answer: d