# Thread: Sum to infinity.. (don't know enough to phrase problem correctly)

1. ## Sum to infinity.. (don't know enough to phrase problem correctly)

This problem is a basic one from finance/economics about obtaining the Present Value; but the math is general
to any field.

How is the value 32,097 obtained?
Is it to do with sum to infinity?

$
PV=20,000.\sum_{i=0}^{infinity}\frac{1}{(1+0.05)^{ 20i}}=32,097
$

I haven't given any context (just a word problem). If needed I will provide.

I need help with this; any help is really appreciated!
thanks

2. Originally Posted by needlittlehelp01
This problem is a basic one from finance/economics about obtaining the Present Value; but the math is general
to any field.

How is the value 32,097 obtained?
Is it to do with sum to infinity?

$
PV=20,000.\sum_{i=0}^{infinity}\frac{1}{(1+0.05)^{ 20i}}=32,097
$

I haven't given any context (just a word problem). If needed I will provide.

I need help with this; any help is really appreciated!
thanks

You have the sum of an infinte geometric series: if $a,ar,ar^2,\ldots,ar^n\ldots$ is an infinite sequence with first element $a$ and common ratio $r$ , and such that $|r|<1$ , then its infinite sum is:

$\sum^\infty_{k=0}ar^k=\frac{a}{1-r}$

Now you first recognize the geometric series, recognize $a\,\,\,and\,\,\,r$ , and then do the infinite sum. You'll check at once that the result indeed is $32,097.0348...$ , which they rounded up to $32,097$.

This stuff is usually learned in high school, and it definitely doesn't belong to abstract algebra.

Tonio

3. Originally Posted by needlittlehelp01
This problem is a basic one from finance/economics about obtaining the Present Value; but the math is general
to any field.

How is the value 32,097 obtained?
Is it to do with sum to infinity?

$
PV=20,000.\sum_{i=0}^{infinity}\frac{1}{(1+0.05)^{ 20i}}=32,097
$

I haven't given any context (just a word problem). If needed I will provide.

I need help with this; any help is really appreciated!
thanks
First, this is in the wrong section- it has nothing to do with "Linear and Abstract Algebra".

Second, that is a "geometric series"- it is of the form $a\sum r^n$ with a= 20000 and $r= \frac{1}{1.05^{20}}= 0.376889$, approximately.

A geometric series has sum $\frac{a}{1- r}$ which, here, is $\frac{20000}{1- 0.376889}= 32097$

Once again, Tonio gets in just ahead of me!

4. Originally Posted by HallsofIvy
...........

Once again, Tonio gets in just ahead of me!

Tonio