# Help with this series please

• March 18th 2011, 03:13 PM
Kuma
Hi can someone help me figure this out:

http://www.webassign.net/cgi-bin/sym...%288%2A%2An%29

I know its convergent, but I'm trying to figure out how to get it into the form:

a/1-r

If you notice, when n = 1, the first term is always 1. Then it becomes 1+7^2/8^2 all the way till n.

I figured that r should be 1/8 since each preceding term differs by a factor of 1/8. Can't figure out a though. Any ideas?
• March 18th 2011, 03:18 PM
tonio
Quote:

Originally Posted by Kuma
Hi can someone help me figure this out:

http://www.webassign.net/cgi-bin/sym...%288%2A%2An%29

I know its convergent, but I'm trying to figure out how to get it into the form:

a/1-r

If you notice, when n = 1, the first term is always 1. Then it becomes 1+7^2/8^2 all the way till n.

I figured that r should be 1/8 since each preceding term differs by a factor of 1/8. Can't figure out a though. Any ideas?

$\displaystyle{\frac{1+7^n}{8^n}<2\left(\frac{7}{8} \right)^n$ , and the rightmost series converges (why?), so the comparison test

yields that also the original series converges.

Tonio
• March 18th 2011, 03:31 PM
Kuma
if i split the series into two parts.

(1/8)^n + (7/8)^n

We can see they both converge since r < 1 for both series.
But how can i figure out what they converge to?
• March 18th 2011, 03:31 PM
Plato
Quote:

Originally Posted by Kuma

Yes is certainly is.
So are $\displaystyle \sum\limits_{k = 1}^\infty {\frac{1}
{{8^k }}} \;\& \,\sum\limits_{k = 1}^\infty {\frac{{7^k }}
{{8^k }}}$

They are rearrangeable, being absolutely convergent.
• March 18th 2011, 05:46 PM
tonio
Quote:

Originally Posted by Kuma
if i split the series into two parts.

(1/8)^n + (7/8)^n

We can see they both converge since r < 1 for both series.
But how can i figure out what they converge to?

The sum of an infinite geometric series with first element $a_1$ and constant

quotient $q\,,\,|q|<1$ , is $\displaystyle{\frac{a_1}{1-q}}$

Tonio