Question

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

Answer 1

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

Answer 2

break it up first

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

r^n = 0 as n->inf

Answer 10

Sure, but to what value does it converge to?

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

if q<1 it will converge!