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Q45E

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Found in: Page 444

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Find the value of $$c$$ if $$\sum\limits_{n = 2}^\infty {{{(1 + c)}^{ - n}} = 2}$$

$$c = \frac{{ - 1 + \sqrt 3 }}{2}$$

See the step by step solution

## Form reduction.

Given that this equation holds:$$\sum\limits_{n = 2}^\infty {{{(1 + c)}^{ - n}} = 2}$$.

This is of the form of a geometric series with ratio $${(1 + c)^{ - 1}}$$

$$\sum\limits_{n = 2}^\infty {{{(1 + c)}^{ - n}} = {{(1 + c)}^{ - 2}}} + {(1 + c)^{ - 3}} + {(1 + c)^{ - 4}} + .... = {\left( {\frac{1}{{1 + c}}} \right)^2} + {\left( {\frac{1}{{1 + c}}} \right)^3} + {\left( {\frac{1}{{1 + c}}} \right)^4} + ...$$

$$= \sum\limits_{n = 1}^\infty {{{\left( {\frac{1}{{1 + c}}} \right)}^2}.{{\left( {\frac{1}{{1 + c}}} \right)}^{n - 1}}}$$

## Comparison with geometric series

After comparison with $$\sum\limits_{n = 1}^\infty {a{r^{n - 1}}}$$, we observe that $$a = {\left( {\frac{1}{{1 + c}}} \right)^2},r = \frac{1}{{1 + c}}$$.

Using geometric formula further, \begin{aligned}\sum\limits_{n = 1}^\infty {a{r^{n - 1}}} & = \frac{a}{{1 - r}} = \sum\limits_{n = 1}^\infty {{{\left( {\frac{1}{{1 + c}}} \right)}^2}.} {\left( {\frac{1}{{1 + c}}} \right)^{n - 1}} = \frac{{{{\left( {\frac{1}{{1 + c}}} \right)}^2}}}{{1 - \frac{1}{{1 + c}}}} = \frac{1}{{{{(1 + c)}^2}}} \times \frac{{(1 + c)}}{c}\\ = \frac{1}{{c + {c^2}}}\end{aligned}

Now we combine this obtained result with $$\sum\limits_{n = 2}^\infty {{{(1 + c)}^{ - n}} = 2}$$(given).

$$2 = \frac{1}{{c + {c^2}}} \Rightarrow 2{c^2} + 2c - 1 = 0$$

Solving using quadratic formula, we obtain:

$$c = \frac{{ - 2 \pm \sqrt {{{(2)}^2} - 4(2)( - 1)} }}{{2(2)}} = \frac{{ - 2 \pm \sqrt {4 + 8} }}{4} = \frac{{ - 2 \pm 2\sqrt 3 }}{4} = \frac{{ - 1 \pm \sqrt 3 }}{2}$$

Now for the given series to be convergent geometric series $$|r| < 1 \Rightarrow |\frac{1}{{1 + c}}| < 1 \Rightarrow |1 + c| > 1$$.

We also see that $$\frac{{ - 1 + \sqrt 3 }}{2} = 0.366{\rm{ and }}\frac{{ - 1 - \sqrt 3 }}{2} = - 1.366$$.

Hence we conclude that $$c = \frac{{ - 1 + \sqrt 3 }}{2}$$since $$|r| < 1 \Rightarrow |\frac{1}{{1 + c}}| < 1 \Rightarrow |1 + c| > 1$$ holds true for this value of $$c$$.

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