I'm not certain how this factor is cancelled out. So one example from an online software I'm using for university gives examples on how to work with a problem, but typically doesn't go through the algebra.  My background in algebra isn't strong, but I'm a bit messed up here.
The problem is
$$
\Sigma^{\infty}_{n=1} {{{(2x-3)}^{2n+1}}\over{n}^{3/2}}
$$
And, I went to use the ratio test.
I'm stuck with it in this form.
$$
\lim_{n\rightarrow\infty}{{{(n)^{3/2}*(2x-3)^{2n+3}}}\over{(n+1)^{3/2}*(2x-3)^{2n+1}}}
$$
And the tutorial shows an identical problem (aside from the 2, being replaced by a 4) factoring out to be 
$$
\lim_{n\rightarrow\infty}(2x-3)^2
$$
Which looks lovely and all, but I'm not certain how the 
$$
n^{3/2}\over{(n+1)}^{3/2}
$$
factors out to be one.  I could simply work with the problem as the example shows but I'd like to know where I goofed.  Thanks for reading my long problem with something that's likely a simple algebra mistake.
 A: It doesn't factor to be $1$, but rather, the quotient is asymptotic to $1$. What that means is they grow at approximately the same rate, and so eventually their ratio is close to $1$.
You can also see that they have the same power, and thus $$ \lim_{n \to \infty} \frac{n^{3/2}}{(n+1)^{3/2}} = 1.$$
And by the way, you're not being asked to simplify with the ratio test. The ratio test is a test for convergence. Please do ensure you're using the jargon correctly.
A: Forget about the exponent and just look at the fraction:
$\lim_{n\rightarrow\infty}$ $n\over{(n+1)}$
This value will never be more than 1, as the denominator will always be greater than the numerator. But if you choose an n value (try n = 10) and then choose another n value (try n = 20) you will notice that whenever you increase the value of n the fraction moves closer to 1.
Once you take the limit to infinity you could think of it as a huge number divided by that same huge number + 1. But if the number was instead small then an increase in 1 might be a large change, however an increase in 1 for a huge number is barely noticeable. So effectively you've got the same number on the top and on the bottom provided the number is large enough. So even if it wasn't +1, and was +3 or +22 or +any integer there would still be a point where n gets so large that the numerator and denominator are also so large they are effectively the same number. This is why the fraction limits to 1.
Now you might feel that it doesn't matter how large of a number, n, is chosen the denominator will always be greater than the numerator. That is correct. Remember that infinity is not a number, it is not a value we can plug into our formula. Infinity is a concept, we tend towards infinity, but we never get there. So this fraction will never be equal to 1, because that would imply the numerator and denominator are equal and that isn't true. But if we were to put a value of n in that was 100 digits long and evaluate the fraction it would be like 0.999999999 or something. We would be happy to just call it 1.
The main point to note here is that n and n+1 are of the same order in terms of n. If we had k, a positive integer, and consider the result of the following limit:
$\lim_{n\rightarrow\infty}$ $n\over{(kn+1)}$ = $1\over{k}$
This is because the numerator and denominator are now of a different order. As n gets very large the denominator will be about k times more than the numerator (k times + 1 to be exact, but +1 doesn't change a big number by much at all).
So long story short you can think of it like this:
$\lim_{n\rightarrow\infty}$ $n\over{(n+1)}$ = $\lim_{n\rightarrow\infty}$ $n\over{n}$ = 1
In your question you have an exponent on the numerator and denominator, but these exponents are the same, so you can think of this as just one exponent operating on the entire fraction. So your problem still reduces to finding $\lim_{n\rightarrow\infty}$ $n\over{(n+1)}$ just remember to raise your answer to the power of 1.5 (in this case it doesn't matter because the answer was 1).
