For the problem
Simplify the following as much as possible without using a calculator. $\sum_{r=-7}^{n} ({\frac{7}{n})^r}, n>0$
I have tried using the geometric sum formula by splitting the summation into two pieces
$\sum_{r=-7}^{0} ({\frac{7}{n})^r}$ + $\sum_{r=1}^{n} ({\frac{7}{n})^r}$
Then by using the formula for geometric series
$\frac{1-r^n}{1-r}$
I came up with the answer:
$\frac{1-(\frac{n}{7})^7}{1-\frac{n}{7}} + \frac{1-(\frac{7}{n})^n}{1-\frac{7}{n}} - 1$
However, the answer does not seem to be right when I check it. Can anyone help me or guide me in the right direction?
Hint: Write it as $ \sum\limits_{s=0}^{n+7} (\frac 7 n)^{s-7}$ (where $s=r+7$) which is $(\frac 7 n)^{-7} \sum\limits_{s=0}^{n+7} (\frac 7 n)^{s}$. Now write down the geometric sum.
You should get $\frac {n^{n+8}-7^{n+8}} {(n-7) 7^{7}n^{n}}$