What Is the Correct Sum of the Series \( \sum_{n=1}^\infty \frac{8^n}{9^n} \)?

[ILoveBaseball]

Determine the sum of the following series

$$\sum_{n=1}^\infty \frac{2^n+6^n}{9^n}$$ or can be written as...
$$\sum_{n=1}^\infty \frac{8^n}{9^n}$$

$$A_1 = 8/9, A_2 = 64/81, A_3 = 512/729$$

common ration (r)= 8/9
first term (a)= 8/9

so plugging everything i know into the geometric series formula:
$$\frac {a}{1-r}$$

i get... 8
but it's wrong, and i don't see why

 

[saltydog]

$$\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$$

And you know what:

$$\frac{k^n}{m^n}$$

is with k and m constants right?

 

[Hurkyl]

Are you sure 2^n + 6^n = 8^n?

 

[ILoveBaseball]

$$\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$$

And you know what:

$$\frac{k^n}{m^n}$$

is with k and m constants right?

$$\frac{2}{9}\sum_{n=1}^\infty \frac{1^n}{1^n}$$ +

$$\frac{6}{9}\sum_{n=1}^\infty \frac{1^n}{1^n}$$


right?

 

[Hurkyl]

I strongly suggest you go back and review the section on exponents in your earlier textbooks -- your problem lies with the fact you don't know how to manipulate them, and it would be much easier to refresh your memory without having to worry about calculus stuff at the same time!

 

[Jameson]

Testing "2" in for x yields the following statements.

$$2^2 + 6^2 = 40$$
$$8^2 = 64$$

Hurkyl is right. Rework that step and see what else you can come up with.

 

[fourier jr]

i got 4. final answer regis
(saltydog did 80% of the work)

 

[ILoveBaseball]

the answer is 2.2857, it was actually an easy question. thanks for the help

 

[Hurkyl]

Really? I find it strange that it happened to be a terminating decimal.

 

[fourier jr]

the answer is 2.2857, it was actually an easy question. thanks for the help

how did you get that? i did $$\frac{1}{1- 2/9} + \frac{1}{1-2/3}$$ which adds up to 4

 

[Hurkyl]

Check the starting index of the sum.

 

[Hippo]

$$\sum_{n=1}^\infty \frac{2^n+6^n}{9^n}$$

So, that's what we're dealing with, right? I'm not too sharp on series, but I don't thing this is hard.

Follow saltydog's advice, and the series will become two series,

$$\sum_{n=1}^\infty \frac{2^n}{9^n} + \sum_{n=1}^\infty \frac{6^n}{9^n}$$

$$\sum_{n=1}^\infty (\frac{2}{9})^n + \sum_{n=1}^\infty (\frac{6}{9})^n$$

Use the geometric series formula on each of those.

i did $$\frac{1}{1- 2/9} + \frac{1}{1-2/3}$$ which adds up to 4

It does?

 

[fourier jr]

sorry 30/7, not 4