Question

in need of a genius for sigma notation?

Answer 1

Identify whether the series \[\sum_{i=1}^{infinite} 12(3/5)^{i-1}\] is a convergent or divergent geometric series and find the sum, if possible.

choices?
This is a convergent geometric series. The sum cannot be found.
This is a divergent geometric series. The sum cannot be found.
This is a divergent geometric series. The sum is 30.
This is a convergent geometric series. The sum is 30.

Answer 2

its the last choice

Answer 3

\[12\frac{1}{1-\frac{3}{5}}\]

right?

Answer 4

Yes, basically. 12/ (1-3/5)

Answer 5

the series doesnt go to zero tho; does that matter as to whether its considered D or C ?

Answer 6

or am i confusing a ration test ....

Answer 7

You're confusing it with the alternating series test. If the limit doesnt go to 0, then it diverges

Answer 8

yes, it is a geometric series

you can prove it is convergent by the ratio test

Answer 9

You can?? Ive never proved a geometric with the ratio test

Answer 10

idk, maybe I'm wrong. I'm out of practice with sequences and series.

Answer 11

actually wolfram alpha says we can:
http://www.wolframalpha.com/input/?i=sum+1+to+infinity+of+12%283%2F5%29%5E%28i-1%29&a=i_Variable

Answer 12

Any infinity series in which the common difference \(r \lt |1|\) is convergent and the sum is given by \(\frac a{1-r}\) where \(a\) is the first term.

Answer 13

Anyone care to help me with my series question??

Answer 14

EDIT: "Any infinite geometric series"

Answer 15

EDIT #2: Common ratio.

Answer 16

Here, since \(\frac 35 <1\) this is a convergent series.

hope this helps.

Answer 17

so that would be using the notation:
\[Sn = \sum_{i}^{infinity} r^i\]
right?

Answer 18

yeah ok

Answer 19

Yes, \(\sum\) is just the summation symbol.

Answer 20

i know I just was confirming which part you meant by "r"