$$ \sum_{n \rightarrow 2}^\infty \frac{(-5)^n}{8^{2n}}$$
I get
$$ \frac{-5}{8^2}(\frac{1}{1-\frac{-5}{8^2}}) $$
but it seems that it is wrong (it is the intermediary step, but I think that this where I am going wrong). I fixed the indices to n-1 and switched the sum expression with $ \frac{a(1-r^n)}{1-r} $ some help?

Question

$$ \sum_{n \rightarrow 2}^\infty \frac{(-5)^n}{8^{2n}}$$
I get
$$ \frac{-5}{8^2}(\frac{1}{1-\frac{-5}{8^2}}) $$ 
but it seems that it is wrong (it is the intermediary step, but I think that this where I am going wrong). I fixed the indices to n-1 and switched the sum expression with $ \frac{a(1-r^n)}{1-r} $ some help?

anonymous · Answer

Ooops the $ n \rightarrow 2 $ should be $n=2$

anonymous · Answer

the equation a (1 - r^n) / (1 - r) tells you that you a is the first term of the sequence, and n is the number of terms

but since n->infinity, and |r| < 1, it becomes a/(1-r)
so what you should have had is:
[(-5)^2/(8^4)] / [1 - (-5)/8^2] = 25/4416

anonymous · Answer

https://www.wolframalpha.com/input/?i=sum+%28-5%29%5En+%2F+8%5E%282n%29%2Cn+%3D+2+to+inf

anonymous · Answer

How did you get the $\frac{-5}{8^2}$ twice on top?

anonymous · Answer

that is when n = 2 (which is the first term of the sequence). 

The index does not tell you what position each term is at, it only tells you the value at that index.

so even though the index starts at 2, the first term is, in this case, (-5)^4/8^4

anonymous · Answer

........ you are right..... cant believe i missed that...... Thanks for the help and time