Solutions - 0150-06

(a) (1) Rewrite general term in standard form for a geometric series:

$$S_N=\sum _{n=0}^N\left(\frac{1}{9}\right){(-\frac{8}{9})}^n$$

Thus $a_0=1/9$ and $r=-8/9$.

(2) Apply “shift method” technique:

Compare $S_N$ and $r{S}_N$:

$$\begin{array}{cc}{S}_N& =\frac{1}{9}+\left(\frac{1}{9}\right){(-\frac{8}{9})}^1+\left(\frac{1}{9}\right){(-\frac{8}{9})}^2+\cdots +\left(\frac{1}{9}\right){(-\frac{8}{9})}^N\\ & \\ (-\frac{8}{9})S_N& =\left(\frac{1}{9}\right){(-\frac{8}{9})}^1+\left(\frac{1}{9}\right){(-\frac{8}{9})}^2+\cdots +\left(\frac{1}{9}\right){(-\frac{8}{9})}^N+\left(\frac{1}{9}\right){(-\frac{8}{9})}^{N+1}\end{array}$$

Subtract and cancel terms:

$$S_N-(-\frac{8}{9})S_N=\frac{1}{9}-\left(\frac{1}{9}\right){(-\frac{8}{9})}^{N+1}$$

Factor and solve:

$$\begin{array}{c}{S}_N(1+\frac{8}{9})=\frac{1}{9}(1-{(-\frac{8}{9})}^{N+1})\\ \\ \gg \gg S_N=\frac{9}{17}\cdot \frac{1}{9}(1-{(-\frac{8}{9})}^{N+1})\\ \\ \gg \gg S_N=\frac{1}{17}(1-{(-\frac{8}{9})}^{N+1})\end{array}$$

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(b) In the limit as $N\to \infty$, this converges to $1/17$ because the term ${(-8/9)}^{N+1}$ converges to $0$.

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(c)

$$S=\frac{1/9}{1--8/9}\gg \gg \frac{1}{17}$$