## College Algebra 7th Edition

convergent; $S_{\infty}=\sqrt2 + 1$
RECALL: (1) The sum of an infinite geometric series is convergent if $|r| < 1$. The sum is given by the formula: $S_{\infty}=\dfrac{a}{1-r}$ (2) An infinite geometric series is divergent if $|r|\ge1$. (3) The common ratio $r$ of a geometric series can be found by dividing any term by the term before it: $r = \dfrac{a_n}{a_{n-1}}$ Solve for $r$ to obtain: $\require{cancel} r = \dfrac{\frac{1}{2}}{\frac{1}{\sqrt2}} \\r=\dfrac{1}{2} \cdot \left(\dfrac{\sqrt2}{1}\right) \\r=\dfrac{\sqrt2}{2}\approx 0.7071$ Since $\left|\dfrac{\sqrt2}{2}\right|<1$, then the series is convergent. Solve for the sum using the formula above, with $a=\dfrac{1}{\sqrt2}$ and $r=\dfrac{\sqrt2}{2}$, to obtain: $\require{cancel} S_{\infty}=\dfrac{\frac{1}{\sqrt2}}{1-\frac{\sqrt2}{2}} \\S_{\infty}=\dfrac{\frac{1}{\sqrt2}}{\frac{2}{2}-\frac{\sqrt2}{2}} \\S_{\infty}=\dfrac{\frac{1}{\sqrt2}}{\frac{2-\sqrt2}{2}} \\S_{\infty} = \dfrac{1}{\sqrt2} \cdot \dfrac{2}{2-\sqrt2} \\S_{\infty}=\dfrac{2}{\sqrt2(2-\sqrt2)} \\S_{\infty}=\dfrac{2}{2\sqrt2-2} \\S_{\infty}=\dfrac{2}{2(\sqrt2-1)} \\S_{\infty}=\dfrac{\cancel{2}}{\cancel{2}(\sqrt2-1)} \\S_{\infty}=\dfrac{1}{(\sqrt2-1)}$ Rationalize the denominator by multiplying $\sqrt2+1$ to both the numerator and the denominator to obtain: $S_{\infty}=\dfrac{1}{(\sqrt2-1)}\cdot \dfrac{\sqrt2+1}{\sqrt2+1}$ Simplify the denominator using the formula $(a-b)(a+b) = a^2-b^2$ to obtain: $S_{\infty}=\dfrac{\sqrt2 + 1}{(\sqrt2)^2-1^2} \\S_{\infty}=\dfrac{\sqrt2 + 1}{2-1} \\S_{\infty}=\dfrac{\sqrt2 + 1}{1} \\S_{\infty}=\sqrt2 + 1$