## Calculus: Early Transcendentals (2nd Edition)

Since the sum has so many terms, the easiest thing to do is write a rule for it. To do that we first need to recognize a pattern. For this one, the terms are being multiplied by $\frac{1}{3}$ each time. That is the $r$ value. The $a$ value is the value of the first term, which is $\frac{1}{4}$ So we have $\Sigma_{n=0}^x\frac{1}{4}(\frac{1}{3})^n$. To find $x$, the last term number, we have to use it's value, which is given to us as $\frac{1}{2916}$ $\frac{1}{4}(\frac{1}{3})^n=\frac{1}{2916}$ $n= log_{\frac{1}{3}}\frac{4}{2916}=6$ $\Sigma_{n=0}^6\frac{1}{4}(\frac{1}{3})^n=\frac{\frac{1}{4}(1-(\frac{1}{3})^7)}{1-(\frac{1}{3})}=.3748$