# Thread: [SOLVED] sum of series

1. ## [SOLVED] sum of series

I am supposed to tell whether this converges or diverges and if it converges tell what it converges to. I'm unclear on how to do the last part.

the summation from n=0 to infinity of 3^(n+1)/4^(n-2)

I used the ratio test and found that it converges. I don't know what to do now.

2. It is equal to $3\cdot4^2 + 3^2 \cdot 4 + \sum_{n=2}^\infty 3^3(\frac34)^{n-2}$.

3. Originally Posted by maddas
It is equal to $3\cdot4^2 + 3^2 \cdot 4 + \sum_{n=2}^\infty 3^3(\frac34)^{n-2}$.
what?

4. I don't know anything about this. And no matter how many different ways I ask this I just get answers that are too complicated for me to understand. I have to figure this out by 9 am. And I don't have a clue as to what I'm doing.

5. Well, write it all out.

$3\cdot 4^2 + 3^2\cdot 4 + 3^3\cdot 1 + {3^4\over 4} + {3^5\over 4^2} + {3^6\over 4^3} +\cdots$

Now factor the later part like

$3\cdot 4^2 + 3^2\cdot 4 + 3^3\Big(1 + {3\over 4} + {3^2\over 4^2} + {3^3\over 4^3} +\cdots\Big)$

The sum in parenthesis is geometric. You should know that 1+r+r^2+...=1/(1-r) for |r|<1. So in this case

$3\cdot 4^2 + 3^2\cdot 4 + 3^3\cdot{1\over 1-\frac34}$

Now what does this expression equal?

Get it? You probably shouldn't wait until the morning before to figure this stuff out :/

6. Originally Posted by maddas
Well, write it all out.

$3\cdot 4^2 + 3^2\cdot 4 + 3^3\cdot 1 + {3^4\over 4} + {3^5\over 4^2} + {3^6\over 4^3} +\cdots$

Now factor the later part like

$3\cdot 4^2 + 3^2\cdot 4 + 3^3\Big(1 + {3\over 4} + {3^2\over 4^2} + {3^3\over 4^3} +\cdots\Big)$

The sum in parenthesis is geometric. You should know that 1+r+r^2+...=1/(1-r) for |r|<1. So in this case

$3\cdot 4^2 + 3^2\cdot 4 + 3^3\cdot{1\over 1-\frac34}$

Now what does this expression equal?

Get it? You probably shouldn't wait until the morning before to figure this stuff out :/
I just got the assignment Fri and I've been trying to figure this out all weekend. And it sorta makes sense. I've never learned about geometric sequences.