# Infinite Series

• Mar 29th 2010, 06:49 PM
stevo25
Infinite Series
I originally thought this Infinite Series would be fairly easy, however I can't figure out how to evaluate it.

infinity
∑ (2^(j+4)) / (7^(j-3))
j=6

is there a certain formula i need to follow?
Thanks.
• Mar 29th 2010, 07:04 PM
Drexel28
Quote:

Originally Posted by stevo25
I originally thought this Infinite Series would be fairly easy, however I can't figure out how to evaluate it.

infinity
∑ (2^(j+4)) / (7^(j-3))
j=6

is there a certain formula i need to follow?
Thanks.

$\sum_{j=6}^{\infty}p^{j+4}{q^{j-3}}=p^4 q^3\sum_{j=6}^{\infty}\left(\frac{p}{q}\right)^j$. Look familiar?
• Mar 29th 2010, 07:08 PM
chiph588@
Quote:

Originally Posted by stevo25
I originally thought this Infinite Series would be fairly easy, however I can't figure out how to evaluate it.

infinity
∑ (2^(j+4)) / (7^(j-3))
j=6

is there a certain formula i need to follow?
Thanks.

$\sum_{j=6}^\infty \frac{2^{j+4}}{7^{j-3}} = 2^4\cdot7^3\sum_{j=6}^\infty \frac{2^{j}}{7^{j}} = 2^4\cdot7^3\sum_{j=6}^\infty\left(\frac27\right)^j$

Now we're left with a geometric series. Can you take it from here?

EDIT: Ah! Drexel28 beat me to the chase!
• Mar 29th 2010, 08:10 PM
stevo25
Ah, it is pretty simple! After that you just add 6 to the j to set the index to 0 and then put (2/7)^6 outside the summation. Then you can plug it into

(ar^(n+1) - a) / (r-1)

formula and take the limit as it approaches infinity.

Thanks!