Solving infinite sequence

[kfoi78]

How is the attached sequence simplified? It seems like a derivative, but I cannot get it to work out algebraically.

 

[MarkFL]

Hello, and welcome to FMH!

The first equation is being subtracted from the second, to obtain the third.

 

[kfoi78]

Hello, and welcome to FMH!

The first equation is being subtracted from the second, to obtain the third.

How do you get the 1 out of the (1+r) ?

And, thanks glad to have found FMH.

 

[MarkFL]

How do you get the 1 out of the (1+r) ?

And, thanks glad to have found FMH.

You can think of distributing:

[MATH](1+r)\times PV=1\times PV+r\times PV=PV+r\times PV[/MATH]

 

[ksdhart2]

Just as a side note, I'm unsatisfied with the given explanation for another reason. They make it seem like this equality holds for all real numbers \(r\), but it really doesn't. It only holds if:

\(\displaystyle \frac{1}{|1 + r|} < 1\)

Or to put it another way:

\(\displaystyle r \in \big( (-\infty, -2) \cup (0, \infty) \big)\)

 

[kfoi78]

Thank you Ksdhart2 for extra clarification. I really do appreciate that.

Thank you Mark FL, but I am still confused as to how you get rid of the PV in the PV+rPV in the result, and how you get rid of the summing sequence?





EDIT: is it beacuse if you minus the PV you can essentially eliminate the infinite sequence, because PV equals the infinite sequence?

I apologize if I am being difficult. But, can I read/study something to better understand the concept at hand here? Is there a geometric representation of this? I am just having a hard time conceptualizing the leap between having (1+r) in the denominator to having just r?

I think I can understand the premise though because 1/3 could be thought of as an infinite series, so certainly a division can result in an infinite series, but I just don't conceptually understand the algebraic leap we took..

 

[MarkFL]

EDIT: is it beacuse if you minus the PV you can essentially eliminate the infinite sequence, because PV equals the infinite sequence?

Yes, that's what I meant by subtracting the first equation from the second.

 

[kfoi78]

Yes, that's what I meant by subtracting the first equation from the second.

That makes a lot more sense now in retrospect thanks! Would you mind reviewing my additional questions in the previous post? If not I understand. I am very thankful for your help!

 

[MarkFL]

A repeating decimal like 1/3 is a good example:

[MATH]S=\frac{3}{10}+\frac{3}{100}+\frac{3}{1000}+\cdots[/MATH]
Now, if we multiply both sides by 10 we get:

[MATH]10S=3+\frac{3}{10}+\frac{3}{100}+\frac{3}{1000}+\cdots=3+S[/MATH]
[MATH]9S=3\implies S=\frac{1}{3}[/MATH]

 

[kfoi78]

Thank you.