What is the value of the summation of (2^n+1)/3^n?

[IntegrateMe]

So the question asks: What is the value of the "summation of" 2ⁿ⁺¹/3ⁿ from "n=1 to infinity."

I changed 2ⁿ⁺¹/3ⁿ into 2*(2/3)ⁿ so i could use it as a geometric series.

So now i just use the rule "a/(1-r) = sum" where a = first term and r = ratio i get 2/(1-(2/3)) which = 6. The answer is 4, any suggestions?

 

[gabbagabbahey]

So now i just use the rule "a/(1-r) = sum" where a = first term and r = ratio i get 2/(1-(2/3)) which = 6. The answer is 4, any suggestions?

Isn't your first term 2/3?

 

[VeeEight]

Your series is going from 1 to infinite. The famous geometric series test is used when the sum starts at 0. So you must subtract the n = 0 term
See here: http://en.wikipedia.org/wiki/Geometric_series#Formula

 

[IntegrateMe]

Oh wait, so the first term is 4/3 so i get (4/3)/[1-(2/3)] = 4. Thank you guys!

 

[gabbagabbahey]

Your series is going from 1 to infinite. The famous geometric series test is used when the sum starts at 0. So you must subtract the n = 0 term
See here: http://en.wikipedia.org/wiki/Geometric_series#Formula

The $a$ in the numerator takes care of that.

$$\sum_{n=n_0}^\infty r^n=\frac{r^{n_0}}{1-r}=\frac{a}{1-r}$$

(When $r<1$ and $n_0\geq0$ of course)