Question about geometric progression

**serenalee** · August 25th 2009, 12:20 AM

Given that sum from n=1 to infinity ( 3^n + 2^n) / 6^n.

Was asked to determine if it's convergent or divergent. If convergent, find its sum.

What i did was

(3/6)^n + (2/6)^n
= (1/2)^n + (1/3)^n

since both 1/2 and 1/3 are less than 1. so it's convergent.
hence I used a/1-r to find the sum to infinity for both 1/2 and 1/3 seperate cases and added up both the sum. But it was wrong. why?

**Prove It** · August 25th 2009, 12:50 AM

Originally Posted by serenalee

Given that sum from n=1 to infinity ( 3^n + 2^n) / 6^n.

Was asked to determine if it's convergent or divergent. If convergent, find its sum.

What i did was

(3/6)^n + (2/6)^n
= (1/2)^n + (1/3)^n

since both 1/2 and 1/3 are less than 1. so it's convergent.
hence I used a/1-r to find the sum to infinity for both 1/2 and 1/3 seperate cases and added up both the sum. But it was wrong. why?

Note that $\frac{3^n + 2^n}{6^n} = \frac{3^n + 2^n}{3^n2^n}= \frac{1}{3^n} + \frac{1}{2^n}$

So $\sum_{i = 1}^{\infty}\left(\frac{3^n + 2^n}{6^n}\right) = \sum_{i = 1}^{\infty}\left(\frac{1}{3^n} + \frac{1}{2^n}\right)$

$= \sum_{i = 1}^{\infty}\frac{1}{3^n} + \sum_{i = 1}^{\infty}\frac{1}{2^n}$

$= \sum_{i = 1}^{\infty}\left(\frac{1}{3}\right)^n + \sum_{i = 1}^{\infty}\left(\frac{1}{2}\right)^n$ .

Each of these is a geometric series. The first has $a = \frac{1}{3}$ and $r = \frac{1}{3}$ . The second has $a = \frac{1}{2}$ and $r = \frac{1}{2}$ .

Since each has $|r| < 1$ , both series are convergent, so the whole series is convergent.

The first series has sum

$S_\infty = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{1}{2}$ .

The second series has sum

$S_\infty = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1$ .

So $\sum_{i = 1}^{\infty}\left(\frac{3^n + 2^n}{6^n}\right) = \frac{1}{2} + 1 = \frac{3}{2}$ .

**HallsofIvy** · August 25th 2009, 08:57 AM

Critical point: notice that the sums, as shown by Prove It: $\sum_{n=1}^\infty \frac{1}{3^n}$ and $\sum_{n=1}^\infty \frac{1}{2^n}$ , start at n= 1. The formula $\sum_{n=0}^\infty ar^n= \frac{a}{1- r}$ has the sum starting at n= 0. Prove It handled that by taking a= 1/3 and 1/2 so they were $\frac{1}{3}\sum_{n=0}^\infty \frac{1}{3}^n$ and $\frac{1}{2}\sum_{n=0}^\infty \frac{1}{2^n}$ .

You could also do that by noting that $\sum_{n=1}^{\infty}r^n$ is just missing the "n= 0" term which is 1. So just use the standard formula, with a= 1, r= 1/3 or 1/2, and subtract 1:
$\frac{1}{1- \frac{1}{3}}- 1= \frac{3}{2}- 1= \frac{1}{2}$ and $\frac{1}{1- \frac{1}{2}}- 1= 2- 1= 1$ .

serenalee, if you got $\frac{3}{2}+ 2= \frac{7}{2}$ instead of $\frac{1}{2}+ 1= \frac{3}{2}$ , that was your error.

(In the future, if you are asking why your answer was wrong, it would be a good idea to tell us what your answer was!)

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