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Expert-verified Found in: Page 443 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # Find the values of x for which the series converges. Find the sum of the series for those values of x.$$\sum\nolimits_{n = 0}^\infty {\frac{{{{(x - 2)}^n}}}{{{3^n}}}}$$

The given series converges for $$- 1 < x < 5$$. Sum of series for these values of $$x = \frac{3}{{5 - x}}$$.

See the step by step solution

## Step 1

Finding the interval where the series converges by using the concept that an infinite geometric whose $$|r| < 1$$ is convergent.

This series converges if we assume it to be of the form of an infinite geometric series, $$\sum\limits_{k = 0}^\infty {a{r^k} = \frac{a}{{1 - r}},|r| < 1}$$

Comparing the given series with the infinite geometric series, we observe that $$a = 1$$ and $$r = \frac{{x - 2}}{3}$$.

Putting |r|<1, $$|\frac{{x - 2}}{3}| < 1$$

ie $$- 1 < \frac{{x - 2}}{3} < 1$$ $$\Rightarrow - 3 < x - 2 < 3 \Rightarrow - 1 < x < 5.$$

Therefore, the series converges when $$x$$ belongs to $$( - 1,5).$$

## Step2

Sum of an infinite geometric progression is given by $$\frac{a}{{1 - r}}$$.

Finding sum of series when $$x$$ is in the range $$( - 1,5).$$

So, we obtain:$$\frac{1}{{1 - (\frac{{x - 2}}{3})}} = \frac{3}{{5 - x}}$$. ### Want to see more solutions like these? 