Find the infinite sum or state that it doesn't exist.

2-3+9/2-27/4+...

On the solutions, it says that r = -3/2 < -1 -->> doesn't exist.

My question is, how did my professor get -3/2 and she gives us a formula: S = a1/1-r , this is my first time learning about infinite sums and i need some help. Thanks! :D

Re: Find the infinite sum or state that it doesn't exist.

r is the common ratio between terms.

The first and second term are 2 and -3, the ratio between them is $\displaystyle \frac{-3}{2}$

The second and third term are -3 and 9/2, the ratio between them is $\displaystyle \frac{\frac{9}{2}}{-3}=\frac{-3}{2}$

Re: Find the infinite sum or state that it doesn't exist.

Quote:

Originally Posted by

**boltage619** 2-3+9/2-27/4+...

On the solutions, it says that r = -3/2 < -1 -->> doesn't exist.

My question is, how did my professor get -3/2 and she gives us a formula: S = a1/1-r , this is my first time learning about infinite sums and i need some help. Thanks! :D

This is a geometric series. What is the *common ratio*?

(By what do we multiply each term to get the next term?) Is it $\displaystyle \frac{-3}{2}~?$

Re: Find the infinite sum or state that it doesn't exist.

Your professor had "insider information"! The simple fact is that seeing just three numbers in a sum does not tell you anything about what the other numbers are! However, typically, we are expected to assume that there is some very simple "rule" governing the numbers. Here those numbers are 2, -3, 9/2, and -27/4 and we notice that 2(-3/2)= -3, -3(-3/2)= 9/2, and (9/2)(-3/2)= -27/4. That is, each number is -3/2 times the previous number. The numbers are 2, 2(-3/2), 2(-3/2)^2, and 2(-3/2)^3. If we **assume** that the series continues in the same way (which is especially reasonable if this is in a textbook chapter on "geometric series"!).

A geometric series is an infinite sum of the form $\displaystyle a+ ar+ ar^2+ \cdot\cdot\cdot+ ar^n+\cdot\cdot\cdot$.

If we look at just a finite sum, $\displaystyle a+ ar+ ar^2+ \cdot\cdot\cdot+ ar^n$ we can show that sums to $\displaystyle \frac{a- ar^{n+1}}{1- r}$. Then, if |r|< 1 taking n higher and higher that "$\displaystyle r^{n+1}$" goes to 0, leaving a limit of $\displaystyle \frac{a}{1- r}$. If |r|> 1, it can be shown that the terms get larger and larger and it does not converge.