Find the sum of the series

1. Aug 19, 2008

greenandblue

1. The problem statement, all variables and given/known data:
The following is a geometric series.
Determine whether series is converges or not.
For the series which converge, enter the sum of the series
$$\sum^{\infty}_{n=1}\frac{7^n+3^n}{8^n}$$

2. The attempt at a solution:
I've looked into calculating $${r}=\frac{a_{n+1}}{a_{n}}$$ but the series isn't constant and neither is r : $$\frac{10}{8}{+}\frac{58}{64}{+}\frac{185}{256}{+...}$$

I feel like there is another approach to solving this problem that I am missing. Your help is appreciated, thanks.

2. Aug 19, 2008

nicksauce

Well it's just the sum of two geometric series isn't it?
$$\sum^{\infty}_{n=1}(\frac{7}{8})^n + \sum^{\infty}_{n=1}(\frac{3}{8})^n$$

3. Aug 20, 2008

greenandblue

I ran those numbers through on paper before too, but it never clicked that I could find the sum individually and add them together. Thanks for your help.

4. Aug 24, 2008

j450n

does it matter that the series is 1-relative? i.e. does one need to convert to make r^(n-1)?

j

5. Aug 24, 2008

HallsofIvy

Good point.
$$\sum_{n=0}^\infty a r^n= \frac{a}{1- r}$$
Here, since it is missing the "n= 0" term, you just subtract that term of: ar0= a so
$$\sum_{n=1}^\infty a r^n= \frac{a}{1- r}- a= \frac{a}{1-r}- \frac{a- ar}{1-r}= \frac{ar}{1- r}$$

Of course, that is exactly the same as if you had just factored out an "r":
[tex]\sum_{n=1}^\infty ar^n= \sum_{n= 1}^\infty (ar)r^{n-1}= \sum_{j= 0}^\infty (ar)r^j[/itex]
where j= n-1 so that when n= 1, j= 1-1= 0.