Convergence of indeterminate forms of a sequence

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 2K views
trap101
Messages
339
Reaction score
0
State whether the sequence converges as n--> ##∞##, if it does find the limit

i'm having trouble with these two:

n!/2n and ∫ e-x2 dx


now I know they're special forms so the ordinary tricks won't work. Any help or hints?
 
Physics news on Phys.org
trap101 said:
State whether the sequence converges as n--> ##∞##, if it does find the limit

i'm having trouble with these two:

n!/2n and ∫ e-x2 dx
For the first, what have you tried?

For the second, that's an integral, not a sequence. How does n approaching infinity enter into things?
trap101 said:
now I know they're special forms so the ordinary tricks won't work. Any help or hints?
 
Mark44 said:
For the first, what have you tried?

For the second, that's an integral, not a sequence. How does n approaching infinity enter into things?


For the first one I simplified it a tad if it's correct to do this:

n!/2n = n (n-1)!/2n = (n-1)!/2 ...so would that tend to ∞?

for the second one:

before being concerned with the integral, e-x2 taking it's limit to ∞ would have the sequnce converge to 0 because e-x2 = 1/ ex2, but shouldn't I integrate it first before I attempt to take the limit?
 
trap101 said:
For the first one I simplified it a tad if it's correct to do this:

n!/2n = n (n-1)!/2n = (n-1)!/2 ...so would that tend to ∞?

for the second one:

before being concerned with the integral, e-x2 taking it's limit to ∞ would have the sequnce converge to 0 because e-x2 = 1/ ex2, but shouldn't I integrate it first before I attempt to take the limit?

For the first one, yes, I think it's pretty clear your simplified form goes to infinity. For the second one you haven't really said how 'n' is involved. Are there limits on your integral? You can't really integrate it in terms of elementary functions. A comparison test might be useful.