Convergent or divergent: SUM 3^-n/2 If it is convergent calculate the sum
its convergent!
it sure is
its a geometric progression!!
\[\sum_{n=0}^{\infty} \]\[3^{-n}/2\]
could you please explain why it it convergent?
cause if we start with n =0 ,1,2 you see that the terms keep getting smaller and smaller.. so its convergent.. only if the terms either remain the same value or keep getting bigger, its a divergent!
ahhh i get it, but what about the sum?
@Mashy that's an awful argument. \[\sum_{n=1}^{\infty} \frac{1}{n}\] also has decreasing terms but is not convergent.
However is true that the sum of a geometric series--- a, ar, ar^2, ar^3, ...---is convergent if and only if |r| < 1. In this case r = 1/3, so the sum does indeed converge.
r= 1/3?!
how can i sole 3^-n / 2 into a standard geometric formula?
james.. 1/n for n = 1 to infinity is A CONVERGENT
At the sum converges to a/(1-r)
@Mashy: bet you $100. Want to take the bet?
lol that would be an absurd bet.. why don't you enlighten me in that case? whats the answer for that???
i know the formula of a/(1-r) but is a 1/2 and what is r?
r = 1/3
i have problems with the \[3^{-n}\]
Just write down the first few term, call them the a_n: a1 = 1/6, a2 = 1/18, a3 = 1/72 ....
@Mashy: We bound below the sum of the series by the sum of another series and show that that lower bound diverges: 1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + 1/10 + ... + 1/16) +... >1 + 1/2 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + (1/16 + .... + 1/16) + ... = 1 + 1/2 + 1/2 + 1/2 + 1/2 + .... It's clear we can write down as many 1/2 terms as we want. Hence this lower sum diverges to +infinity and hence the original sum does as well.
So it's clear that a1 = a a2 = ar a3 = ar^2 where a = 1/6 and r = 1/3 = 3^(-1)
wow james... i take back.. glad i didn't bet lol.. i learnt something today :D thank you!!!!
I wish you had. $100 would have come in handy. Next time.
;-)
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