Question

Please help! Find the following infinite sum, if it exists. If one does not exist enter "no infinite sum".
(LOOK AT RESPONSE)

Answer 1

\[\sum_{n=1}^{\infty}6\left( 2/3 \right)^{n-1}\]

Answer 2

is it n^(-1) or is it n-1? :)

Answer 3

n-1

Answer 4

\[\sum_{n =1}^{\infty} 6(\frac {2}{3})^{n-1}\]

Answer 5

What is going on with Latex? That looks like crap

Answer 6

Yes Hero, that is correct. My bad :/

Answer 7

alright, I think I got it :\[\sum_{n=1}^{\infty} 6(\frac{2}{3})^{n-1} = \sum_{n=1}^{\infty} 2^n\]

then you can use the root test to proceed ^_^ got it?

Answer 8

since you'll be left with :\[2(2)^{n-1} = 2^{n-1+1} = 2^n\]

right? .-.

Answer 9

or is that wrong? hmmm

Answer 10

hero what do you think? ^_^

Answer 11

I'm not sure... Hero?

Answer 12

Let me see if I can verify that...

Answer 13

I get 18 for the sum

Answer 14

meh, then that means my steps were wrong ^_^"?

Answer 15

Hmm. I'll try. Thank you for the help guys :)

Answer 16

np and good luck ! ^_^

Answer 17

18 is correct - I usually find it easiest to look at the first few terms, and from there figure out what the first term (a) and the common ratio (r) are. In this case we have:

6, 4, 8/3, 16/9 etc..., so a=6 and r=2/3. Then use the geometric series formula:

Sum = a/(1-r) = 6/(1-2/3) = 18