Solving Geometric Series: Find x Values |r|<1

• Briggs
In summary, Daniel is having trouble understanding geometric series. Geometric series is a math concept where you take a certain number and add it to a previous number to get a new number. For example, in the conversation, Daniel is trying to figure out whether the series 5+25(3x+4)+125(3x+4)^2 converges or not. The first series is easy because the equation is 2+4x+8x^2+16x^3. However, the equation for the second series becomes 5(3x+4) which doesn't converge because r=5(3x+4) is not a valid equation for a geometric series. However, by understanding the modulus
Briggs
I am having a little trouble with some questions on geometric series'
For example, Find the values of x for which the following geometric series converge
I have done the first one easy enough
$$2+4x+8x^2+16x^3...$$
$$r=2x$$
$$|r|<1$$
$$|x|<\frac{1}_{2}$$
$$\frac{-1}{2} < x < \frac{1}_{2}$$

But then it gets more difficult and adds more to the question for example
$$5+25(3x+4)+125(3x+4)^2$$
I get $$r=5(3x+4)$$
so $$|5(3x+4)|<1$$

But then I am stuck as to what to do next any hints on how to go about these questions would be appreciated.

It's the same kind of story to explicitate the modulus...

$$\left|a\right|<b\Leftrightarrow -b<a<b$$

Thankfully,the ratio is linear in "x"...

Daniel.

EDITalls,this is the kind of terminology i had been used to in high school.Always the teacher said about "modulus explicitation" (sic!)...

Last edited:
"explicate the modulus"?? Wow! Them big words is too much for me!

Briggs, what dextercioby means is: |5(3x+4)|< 1 is exactly the same as

-1< 5(3x+4)< 1. Now solve for x.

HallsofIvy said:
"explicate the modulus"??

It's explicitate!

Thanks for the help, it seems quite simple now that you have shown me the next step, I get

$$-1<15x+20<1$$

$$\frac{-1-20}{15}<x<\frac{1-20}{15}$$

$$\frac{-7}{5}<x<\frac{-19}{15}$$

Which is the correct answer so thanks a lot for the help

To help vizualize this geometrically, imagine a point P on the number line. Then, all points X that are strictly inside a distance d from the fixed point P, are given by the equation : |X-P| < d

In other words, this means that X is the set of points in (P-d, P+d). This follows directly from the definition of the modulus :

Definition : |x| = x if x > 0 and |x| = -x if x <= 0

Derivation of Geometric Statement : |X-P| = X-P if X > P and |X-P| = P - X, otherwise (from def'n.)

In the first case (when X > P), |X-P| < d means that X-P < d or X < P+d. So : P < X < P+d

In the second case (when X <= P) |X-P| < d means that P-X < d or X > P-d. So : P-d < X <= P

Combining the above two cases, you see that |X-P| < d gives P-d < X < P+d.

Example : |ax + b| < d is the same as writing |ax - (-b)| < d, which translates as "the distance of points ax from the point b is less than d. So, b-d < ax < b+d, or (b-d)/a < x < (b+d)/a

1. What is a geometric series?

A geometric series is a sequence of numbers where each term is multiplied by a constant ratio to get the next term. For example, in the series 2, 6, 18, 54, the ratio between each term is 3.

2. How do I find the sum of a geometric series?

The formula for finding the sum of a finite geometric series is S = a(1-r^n)/(1-r), where a is the first term, r is the common ratio, and n is the number of terms. If the series is infinite, the sum is equal to a/(1-r).

3. What is the condition for convergence in a geometric series?

A geometric series will only converge if the absolute value of the common ratio (r) is less than 1. This means that the terms in the series will get smaller and smaller as n approaches infinity, resulting in a finite sum.

4. How do I find the value of x in a geometric series?

To find the value of x in a geometric series, you must first determine if the series is convergent. If it is, then you can use the formula x = a/(1-r) to solve for x, where a is the first term and r is the common ratio. If the series is not convergent, then there is no value of x.

5. Can a geometric series have a negative common ratio?

Yes, a geometric series can have a negative common ratio. The important condition for convergence is that the absolute value of the ratio is less than 1. For example, the series -6, 3, -1.5, 0.75 is a geometric series with a common ratio of -0.5 and it converges.

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