# Determine Convergence or Divergence. If conv. find the sum:

1. May 5, 2012

### knowLittle

1. The problem statement, all variables and given/known data
$\sum \dfrac {1+2^{n}} {3^{n}}$

According to Wolfram Alpha the sum is 5/2. But, I think that my method is fine and shows another result.

3. The attempt at a solution
$\sum \dfrac {1+2^{n}} {3^{n}}=\sum \left[ \left( \dfrac {1} {3}\right) ^{n}+\left( \dfrac {2} {3}\right) ^{n}\right]$

Now, by geometric series property and property of infinite series:
$\dfrac {1} {1-\dfrac {1} {3}}+\dfrac {1} {1-\dfrac {2} {3}}=\dfrac {3} {2}+3$ =9/2 =4.5

Also, I found by the ratio test that it is absolutely convergent. (2/3) < 1

I don't know, why Wolfram Alpha shows that the final sum is 5/2.

Thank you.

Last edited: May 5, 2012
2. May 5, 2012

### LCKurtz

You didn't show where the index of summation starts. Is it $n=1$ or $n=0\$? It affects the first term and the sum.

3. May 5, 2012

### knowLittle

Sorry, the sum goes from n=1 to infinity.

4. May 5, 2012

5. May 5, 2012

### knowLittle

Ok. I still don't see it.
S1= 3/3 =1
S2= 5/9
S3=9/27
S4=17/81

I just checked my textbook and Geometric series start from n=0. This is what you are referring to.
Ok, so I cannot use the Geometric series method. How can I find a solution?

6. May 5, 2012

### LCKurtz

If the first term of a geometric series is $a$ and the common ration is $r$ then the sum of the series is $\frac a {1-r}$. If you phrase it that way, it doesn't matter how it is indexed.

7. May 5, 2012

### knowLittle

Ok. So, what you are saying is that my 1st term is equal to a. Then, a=1. I can't determine the ratio from the summation. Could you help?

I want to correct my previous statement for the sums:
S1=1
S2=1+5/9=14/9
S3=(9/27)+14/9=51/27
S4=17/81 +51/27 = 170/80
.
.
. so on

8. May 5, 2012

### Bohrok

Since the index starts at 1, that means the first terms of the series are 1/3 and 2/3 respectively. And with a representing the first term of each from what LCKurtz said,$$\sum^\infty_{n=1} \left[ \left( \dfrac {1} {3}\right) ^{n}+\left( \dfrac {2} {3}\right) ^{n}\right] = \dfrac {\frac{1}{3}} {1-\dfrac {1} {3}}+\dfrac {\frac{2}{3}} {1-\dfrac {2} {3}}$$

9. May 5, 2012

### knowLittle

Wow, so in this case. a is not only the coefficient, but also the ratio.

Ok, thank you. I like to understand.

10. May 5, 2012

### Ray Vickson

Of course you can use the geometric series method; you just have to subtract the n=0 terms from what you wrote. That will give you the sum from n=1 to ∞. Alternatively, you can recognize that for |r| < 1 you have $r + r^2 + r^3 + \cdots = r(1 +r + r^2 + \cdots),$ and use your geometric sum result for the quantity in brackets.

RGV