Thread: Proof of a limit of an infinite sequence

Find and prove

I'm thinking of rationalizing it first, also I can see that the limit approaches 0

That sounds good- multiply by

yes and it becomes

From


You can show that this is always less than


and always greater than


Using the sandwich theorem show that a_n converges to zero.

Originally Posted by Shakarri

From


You can show that this is always less than


and always greater than


Using the sandwich theorem show that a_n converges to zero.

Isn't the sandwich theorem overkill here? The denominator clearly goes to infinity, so the whole fraction goes to 0.

Oh yes you could do it that way

right but when proving, for all epsilon greater than 0, what N would I choose so that
If i try to square it, epsilon gets squared too...

Originally Posted by Kanwar245

right but when proving, for all epsilon greater than 0, what N would I choose so that |3/2sqrt(n)| < epsilon
If i try to square it, epsilon gets squared too...

Why bother squaring anything? Since this is a sequence, we know n is positive, and so the stuff inside the absolute value has to be positive. This means its equal to its absolute value (or the absolute value sign can be neglected).

Right so we have this:



Now from this point on, if I try to isolate n, at some point I will have to square which will yield ...

Originally Posted by Kanwar245

Right so we have this:



Now from this point on, if I try to isolate n, at some point I will have to square which will yield ...

What's wrong with that?

Originally Posted by Prove It

What's wrong with that?

Oh so we can do this?




So when proving, I can say
Given
Choose ?

Originally Posted by Kanwar245

Oh so we can do this?




So when proving, I can say
Given
Choose ?

You certainly can

Okay Perfect!