No friends the limit comparison test doesn't go like that.
Also, the limit of the sequence is 1, not 0, so by Divergence Test the series diverges.
Does it converge or diverge? If it converges, then what does it converge to?
Series with n = 1 to infinity
This one would use the limit test, but I can't find (The series to compare it to). Any strategies for finding it?
Limit Test is: Limit as n goes toward infinity of
Rules:
If is infinity then diverges
If then converges
If comes out to a value, then if converges, then converges, and if diverges, then diverges.
Limit comparison test is like this:
Given a(n), comparing it to b(n), both of which are positive.
The test is valid only if lim(n-> inf.) a(n)/b(n) gives you a non-zero finite value: b(n) converges then a(n) also converges; b(n) diverges then a(n) also diverges.
One thing you have to do is find out what should be. Therefore, you have to simplify the n terms. Correct? How would we go about this? The correct should be How do we get there using the arithmetic of exponents?
How is simplification and exponential arithmetic done on
??
NO! You are supposed to know that a series only has a chance of converging if the terms of the series vanish to 0. So if the limit of your terms is NOT 0, then the series can not converge!
$\displaystyle \begin{align*} \frac{n^3 - 3}{n^3 + n^2 + 1} &= \frac{n^3 + n^2 + 1 - n^2 - 4}{n^3 + n^2 + 1} \\ &= \frac{n^3 + n^2 + 1}{n^3 + n^2 + 1} - \frac{n^2 + 4}{n^3 + n^2 + 1} \\ &= 1 - \frac{n^2 + 4}{n^3 + n^2 + 1} \\ &= 1 - \frac{\frac{1}{n} + \frac{4}{n^3}}{1 + \frac{1}{n} + \frac{1}{n^3}} \\ &\to 1 - 0 \textrm{ as } n \to \infty \\ &= 1 \end{align*}$
The terms do NOT approach 0, so the series diverges!