Question
Download Solution PDFThe sum of the series \(3 - 1 + \frac{1}{3} - \frac{1}{9} + \ldots \) is equal to
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
If a, ar, ar2, …. is an infinite GP, then the sum of infinite geometric series is given by.
Sum of infinite GP = \({s_\infty } = \;\frac{a}{{1\; - \;r}}\); |r| < 1
Calculation:
Here, we have to find the sum of the series \(3 - 1 + \frac{1}{3} - \frac{1}{9} + \ldots \)
As, we can see that sequence is an GP with a = 3 and r = - 1/3
Sum of the series = S∞ = a/ (1 - r)
Sum of the series = \(\frac{3}{{1\; - \;\frac{{ - 1}}{3}}} = \;\frac{3}{{1 + \;\frac{1}{3}}} = \;\frac{3}{{\left( {\frac{4}{3}} \right)}} = \;\frac{9}{4}\)Last updated on Jul 8, 2025
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