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Sequence and Series

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Question : 20 of 86

Marks: +1, -0

The sum of the series

1,\;\frac{1}{3},\;\frac{1}{3^{2}},\;\frac{1}{3^{3}},\;\dots

\infty

is

$\;\frac{4}{3}$
$\;\frac{3}{2}$
$\;\frac{1}{3}$
None

Solution:

Concept:

This is an infinite geometric progression (GP) with first term

a = 1

and common ratio

r = \frac{1}{3}

. Since

|r| < 1

, the sum to infinity is given by

S_\infty = \frac{a}{1 - r}

.

Explanation:

The series is

1, \frac{1}{3}, \frac{1}{3^2}, \frac{1}{3^3}, \dots

Here,

a = 1

and

r = \frac{1}{3}

.

Using the formula:

S_\infty = \frac{a}{1 - r} = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2}

.

Answer:

\frac{3}{2}

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