Summations and Geometric Series

Since $\sum_{n=0}^{\infty} \frac{1}{k^n} = \alpha \cdot \sum_{n=1}^{\infty} \frac{1}{k^n},$ $\alpha$ must equal $\frac{\sum_{n=0}^{\infty} \frac{1}{k^n}}{\sum_{n=1}^{\infty} \frac{1}{k^n}}.$ So, to find $\alpha$ we must simply find the values of $\sum_{n=1}^{\infty} \frac{1}{k^n}$ and $\sum_{n=0}^{\infty} \frac{1}{k^n}$ , and then divide. We can find the value of $\sum_{n=1}^{\infty} \frac{1}{k^n}$ by using the geometric series formula. Expanding $\sum_{n=1}^{\infty} \frac{1}{k^n}$ gives: $\frac{1}{k}+\frac{1}{k^2} + \frac{1}{k^3} + \frac{1}{k^4} \dots =\frac{\frac{1}{k}}{1-\frac{1}{k}} = \frac{\frac{1}{k}}{\frac{k}{k} - \frac{1}{k}} = \frac{\frac{1}{k}}{\frac{k-1}{k}} = \frac{1}{k} \cdot \frac{k}{k-1} = \boxed{\frac{1}{k-1}}.$ We can use the value of $\sum_{n=1}^{\infty} \frac{1}{k^n}$ (above) to then find the value of $\sum_{n=0}^{\infty} \frac{1}{k^n}$ : $\sum_{n=0}^{\infty} \frac{1}{k^n} = \frac{1}{k^0} + \sum_{n=1}^{\infty} \frac{1}{k^n} = \frac{1}{1}+\frac{1}{k-1} = \frac{k-1}{k-1}+\frac{1}{k-1} = \frac{k-1+1}{k-1} =\frac{k}{k-1}.$ Since we now know the values of $\sum_{n=1}^{\infty} \frac{1}{k^n}$ $\left( \frac{1}{k-1} \right)$ and $\sum_{n=0}^{\infty} \frac{1}{k^n}$ $\left( \frac{k}{k-1} \right)$ , we must now simply divide to find $\alpha$ : $\alpha=\frac{\sum_{n=0}^{\infty} \frac{1}{k^n}}{\sum_{n=1}^{\infty} \frac{1}{k^n}} = \frac{\frac{k}{k-1}}{\frac{1}{k-1}}=\frac{k}{k-1} \cdot \frac{k-1}{1} = \boxed{k}$ So the value of $\alpha$ is $\boxed{k}$ .

No where is it mentioned except by implication, that k must be > 1. If we let alpha = k, and multiply it into the series on the right we have a sum from n = 1 to infinity of 1/[k^(n - 1)} Letting m = n -1, we have the sum from m = 0 to infinity of 1/[(k^m)], agreeing with the sum on the left. Ed Gray