Question

prove that the sequence 2^n diverges

Answer 1

can you do this by contradiction?

Answer 2

|x_n - L|>= epsilon
needs to be this form

Answer 3

I just dont know how to show for L<0

Answer 4

If we assume the sequence converges instead, then this implies:\[\lim_{k\rightarrow\infty}2^k=0\]but we know this is not true, therefore it diverges.

Answer 5

i cant use limits in proof, needs to be formal proof

Answer 6

ok - beyond my realm then. hopefully someone else can help out. sorry :)

Answer 7

np ty

Answer 8

yw :)

Answer 9

for L<0
let N be in natural and let n be some n>N and let epsilon = 1
then |x_n-L| = x_n+L > 1 = epsilon
this is the case for L<0 but I dont know how to do it for L>=0

Answer 10

for L>0 \[\let \[nn=\max(N,\left[ \frac{\ln("\epsilon=1"+L)}{\ln(2)} \right]+1,\left[ \frac{\ln(L)}{\ln(2)} \right]+1)\]
n>=N and n>ln(L)/ln(2)====>2^n>L====>|x_n-L|=2^n-L
and n>ln(1+L)/ln(2)=====> 2^n>1+L===> 2^n-L>1
now the case l=0 you put it with the case L<0 wich is also correct for L=0

Answer 11

for these methods ! we "usually" use the backwards manner !(to find "n" for divergence;and "N" for convergence) and any questions are welcome @zzr0ck3r :) !

Answer 12

great I got to think about that, we never use the max thing, so trying to figure it out.

Answer 13

ohh ! It makes things a lot easier :) ! ( at least for me ) !
Just in case you didn't understand something...I'm ready !
"max" is the greatest of those numbers ! I like to note it like that !