"Find all values of x for which the..." solved and explained

beljuA

Answered question

2021-02-25

Find all values of x for which the series converges. For these values of x, write the sum of the series as a function of x.

\sum_{n = 1}^{\infty} (3 x)^{n}

Answer & Explanation

tabuordg

Skilled2021-02-26Added 99 answers

Consider the series
$\sum_{n = 1}^{\infty} (3 x)^{n}$
Consider a G.P series $\sum_{n = 1}^{\infty} r^{n}$
Series is convergent for $| r | < 1$
Sum of this infinite series is given by
$S_{\infty} = \frac{a}{1 - r}$
$\sum_{n = 1}^{\infty} (3 x)^{n} = (3 x)^{1} + (3 x)^{2} + (3 x)^{3} + (3 x)^{4} . . . .$
$\sum_{n = 1}^{\infty} (3 x)^{n} = (3 x) + (3 x)^{2} + (3 x)^{3} + (3 x)^{4} . . . .$
The given series is a Geometric progression with the
first term $= 3 x$
common ratio $= 3 x$
The given series is convergent for

$| 3 x | < 1$

$\Rightarrow - 1 < 3 x < 1$

$\Rightarrow - \frac{1}{3}$

Sum of the series

$\sum_{n = 1}^{\infty} (3 x)^{n}$

for $- \frac{1}{3}$

Here

$a = 3 x, r = 3 x$

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

$S_{\infty} = \frac{3 x}{1 - 3 x}$

Answer:

Series is convergent for $- \frac{1}{3}$

Sum of the series is $\frac{3 x}{1 - 3 x}$

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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