# Thread: Alternating sequence, convergence and its limit

1. ## Alternating sequence, convergence and its limit

The question asks to verify whether it converges and what the limit is if it does

This is the sequence where n ∈ N

(n^4 − 3n^2 + 1)/(n− √[1 + 4n^8])

I began attacking this via the sandwich theorem, but then realised that the square root in the denominator means there are two answers for each value of n... therefore is it even an alternating sequence?

outside the brackets of the sequence there is an ∞ at the top and n=1 at the bottom
... does this mean I should only consider the positive values?

thanks

2. Originally Posted by richmond91
The question asks to verify whether it converges and what the limit is if it does

This is the sequence where n ∈ N

(n^4 − 3n^2 + 1)/(n− √[1 + 4n^8])

I began attacking this via the sandwich theorem, but then realised that the square root in the denominator means there are two answers for each value of n... therefore is it even an alternating sequence?

outside the brackets of the sequence there is an ∞ at the top and n=1 at the bottom
... does this mean I should only consider the positive values?

thanks
Divide top and bottom by $n^4$. Proceed from there.

3. Originally Posted by richmond91
The question asks to verify whether it converges and what the limit is if it does

This is the sequence where n ∈ N

(n^4 − 3n^2 + 1)/(n− √[1 + 4n^8])

I began attacking this via the sandwich theorem, but then realised that the square root in the denominator means there are two answers for each value of n... therefore is it even an alternating sequence?
No, the square root does NOT mean that. $\sqrt{a}$ is the positive number satisfying $x^2= a$.

outside the brackets of the sequence there is an ∞ at the top and n=1 at the bottom
... does this mean I should only consider the positive values?

thanks
Yes, n can be any positive integer.

4. cool... I think I cracked it.. After doing my proof I arrive at the convergence at -0.5