Series: Find the sum of 4/3 + 6/9 + 8/27 + 10/81 + ....

[Math wiz ya rite 09]

Ok. Here's my question:

Find the sum:

S= 4/3 + 6/9 + 8/27 + 10/81 + 12/243 + 14/729 + ...

I believe this is known as an infinite series, but not sure on the whole terminology.

Could someone please show me the steps towards the completeion of this problem.

THANKS!

 

[stapel]

The first step would be to figure out a formula for the n-th term. I would suggest looking at the numerators and denominators separately:

. . . . .4, 6, 8, 10, 12,...

. . . . .3, 9, 27, 81, 243,...

Can you get anywhere with that?

Eliz.

 

[galactus]

Here are some hints. Identities if you will:

\(\displaystyle \L\\\sum_{n=1}^{\infty}\frac{n}{a^{n}}=\frac{a}{(a-1)^{2}}\)

\(\displaystyle \L\\\sum_{n=1}^{\infty}\frac{1}{a^{n}}=\frac{1}{a-1}\)

 

[TchrWill]

Find the sum:

S= 4/3 + 6/9 + 8/27 + 10/81 + 12/243 + 14/729 + ...

The limit of (4 + 2(n - 1))/3^n = ?

 

[soroban]

Hello, Math wiz ya rite 09!

Find the sum: \(\displaystyle \:S\;=\; \frac{4}{3}\,+\,\frac{6}{9}\,+\,\frac{8}{27}\,+\,\frac{10}{81}\,+\,\frac{12}{243}\,+\,\frac{14}{729}\,\cdots\)

This is an infinite series ... unfortunately, not a geometric series.
. . but we can create one.


. . .We have: \(\displaystyle \L\;\;S\;=\; \frac{4}{3}\,+\,\frac{6}{9}\,+\,\frac{8}{27}\,+\,\frac{10}{81}\,+\,\frac{12}{243}\,+\,\frac{14}{729}\,\cdots\)

Multiply by \(\displaystyle \frac{1}{3}:\L\;\;\frac{1}{3}S\;=\;\;\;\;\;\frac{4}{9}\,+\,\frac{6}{27}\,+\,\frac{8}{81}\,+\,\frac{10}{243}\,+\,\frac{12}{729}\,+\,\cdots\)

. . Subtract: \(\displaystyle \L\;\frac{2}{3}S\;=\;\frac{4}{3}\;+\;\underbrace{\frac{2}{9}\,+\,\frac{2}{27}\,+\,\frac{2}{81}\,+\,\frac{2}{243}\,+\,\frac{2}{729}\,+\,\cdots}_{\text{a geometric series}}\)

The geometric series has first term \(\displaystyle a\,=\,\frac{2}{9}\) and common ratio \(\displaystyle r\,=\,\frac{1}{3}\)
. . Its sum is: \(\displaystyle \L\,\frac{\frac{2}{9}}{1\,-\,\frac{1}{3}} \:=\:\frac{1}{3}\)

So we have: \(\displaystyle \L\:\frac{2}{3}S\;=\;\frac{4}{3}\,+\,\frac{1}{3}\;=\;\frac{5}{3}\)

Therefore: \(\displaystyle \L\:S\;=\;\frac{5}{2}\)