TAOCP 1.2.3 Exercise 12
The given sum is a finite geometric progression with first term $a = 1$, common ratio $x = \frac{1}{7}$, and $n+1$ terms.
Section 1.2.3: Sums and Products
Exercise 12. [10] What is $1 + \frac{1}{7} + \frac{1}{49} + \frac{1}{343} + \cdots + \left(\frac{1}{7}\right)^n$?
Verified: yes
Solve time: 36s
The given sum is a finite geometric progression with first term $a = 1$, common ratio $x = \frac{1}{7}$, and $n+1$ terms. By equation (14), the sum of a geometric progression is
$\sum_{0 \le j \le n} a x^j = a \frac{1 - x^{n+1}}{1 - x}.$
Substituting $a = 1$ and $x = \frac{1}{7}$, we obtain
$\sum_{j=0}^n \left(\frac{1}{7}\right)^j = \frac{1 - \left(\frac{1}{7}\right)^{n+1}}{1 - \frac{1}{7}} = \frac{1 - \left(\frac{1}{7}\right)^{n+1}}{\frac{6}{7}} = \frac{7}{6} \left( 1 - \frac{1}{7^{,n+1}} \right).$
Hence, the sum is
$\boxed{\frac{7}{6} \left( 1 - \frac{1}{7^{,n+1}} \right)}.$