OpenStudy (kainui):

$\sum_{k=0}^{\infty}[3(2/5)^k-2(5/7)^k]$ How would I go about evaluating this infinite series?

5 years ago
OpenStudy (anonymous):

yes, you should check and see if its geometric, which it looks like it might be

5 years ago
OpenStudy (anonymous):

break it up first

5 years ago
OpenStudy (experimentx):

since r is less than one ... i guess it would be convergent.

5 years ago
OpenStudy (kainui):

Sure, I can break it up real fast for you, but after that what happens?

5 years ago
OpenStudy (anonymous):

$3\sum(\frac{2}{5})^k-2\sum(\frac{5}{7})^k$ for each use $\frac{a}{1-r}$

5 years ago
OpenStudy (kainui):

$3\sum_{k=0}^{\infty}(2/5)^k -2 \sum_{k=0}^{\infty}(5/7)^k$

5 years ago
OpenStudy (anonymous):

actally in this case $$a=1$$ so use $$\frac{1}{1-r}$$

5 years ago
OpenStudy (kainui):

Why am I using that arbitrary formula a/(1-r)?

5 years ago
OpenStudy (experimentx):

r^n = 0 as n->inf

5 years ago
OpenStudy (kainui):

Sure, but to what value does it converge to?

5 years ago
OpenStudy (experimentx):

3(1/ (1 -2/5)) - 2(1/(1 - 5/7))

5 years ago
OpenStudy (kainui):

I'm not following you. How does it converge to this?

5 years ago
OpenStudy (experimentx):

use geometric series formula Sn = a(1-r^n)/(1-r)

5 years ago
OpenStudy (anonymous):

if $$0<r<1$$ we get $$a+ar+ar^2+ar^3+...=\sum_{k=0}^{\infty}ar^k=\frac{a}{1-r}$$

5 years ago
OpenStudy (anonymous):

i would hardly call the formula arbitrary. you can find its derivation in any book on the subject, but it is how one sums a geometric series, which is what you need for this problem

5 years ago
OpenStudy (anonymous):

GP: $Sn=a1((q^{n}-1/(q-1) )$ since q<1 when n->infinity $Sn=a1*(1/(1-q))$ as satelitte said

5 years ago
OpenStudy (kainui):

Can someone show me how to derive the geometric infinite series formula, I can't seem to find it or figure it out.

5 years ago
OpenStudy (anonymous):

do you know the finite formula of sum?just take the limi when n->inf

5 years ago
OpenStudy (anonymous):

if q<1 it will converge!

5 years ago