infinite series

[mark]

detrermine whether the series converges,and if so find its sum

sigma (2/3)^k+2
(k = 1 to
infinity)

i used the geometric series formula

sigma ar^k = a/1-r
(k=0 to
infinity)

i choose a = 1, r = 2/3

1/(1 -2/3)= 4/3

i dont know what i did wrong

 

[daon]

\(\displaystyle \sum (\frac{2}{3})^{k+2}\) or \(\displaystyle \sum [(\frac{2}{3})^{k} + 2]\) ?

I am assuming it is the former, as the latter does not converge.

Then change the sum \(\displaystyle \sum (\frac{2}{3})^{k+2}\) into:

\(\displaystyle \sum (\frac{2}{3})^2 * (\frac{2}{3})^{k}\)

= \(\displaystyle \frac{4}{9}\sum (\frac{2}{3})^{k}\)

Also, your a = \(\displaystyle \frac{2}{3}\) not 1 (since your summation starts at 1 and not 0, so:

Your sum will be \(\displaystyle \frac{4}{9} * \frac{\frac{2}{3}}{1-\frac{2}{3}}\)

Daon

 

[soroban]

Hello, mark!

Determine whether the series converges, and if so find its sum
\(\displaystyle \L\;\;\;\sum^{\infty}_{k=1}\left(\frac{2}{3}\right)^{k+2}\)

I used the geometric series formula: \(\displaystyle \L\,\sum^{\infty}_{k=0}ar^k\;=\;\frac{a}{1\,-\,r}\;\) . . . yes!

I chose: \(\displaystyle \,\underbrace{a\, =\, 1},\;\; r\, =\,\frac{2}{3}\;\)
. . . . . . . . no

Take a look: \(\displaystyle \L\;\sum^{\infty}_{k=1}\left(\frac{2}{3}\right)^{k+2}\;=\;\left(\frac{2}{3}\right)^3\,+\,\left(\frac{2}{3}\right)^4\,+\,\left(\frac{2}{3}\right)^4\,+\,\cdots\)

The first term is: \(\displaystyle \,a\,=\,\frac{8}{27}\)