Some special tests for convergence. Please, I need some explanations on some special tests for convergence of a series.
For example, the ratio test, comparison test and root test. 
Ratio test: let the limit as $n$ tends to $\infty$ of $\frac{|U_{n+1}|}{|U_{n}|}=L$, now if $L<1$, the series converges and if $L>1$, it diverges. If $L=1$, The test is inconclusive.
Comparison test: Is also a test for convergence of a series by considering a series that converges already in such a way that $U_n\le V_n $. Where $\sum U_n$ is the series that we are to test its convergence while $\sum V_n$ is a convergent series already.  
Root Test: let the limit as n tends to $\infty$ of $\sqrt[n]{|U_{n}|}=L$, now if $L<1$, the series converges and if $L>1$, it diverges. If L=1, The test is inconclusive. 

What is confusing me is when to make use of this test when a series is given, I mean which one to use. 
  For example, consider the series $\frac{n}{(2n+1)(2n-1)}$. Here, we can use the comparism test and also the ratio test but one diverges and  while the other test fails. So, in this case and some other cases too, how am I going to know the series required?

 A: It is not easy to give an effective answer but here are some ideas.
1) The ultimate test is comparison. All others are particular cases of it. So, in terms of power this is the one. But it is not easy to guess with what to compare.
2) The ratio test (in the form you wrote them) is slightly more powerful than the root test. This is because $\lim \frac{a_{n+1}}{a_n}=L$ implies $\lim \sqrt[n]{a_n}=L$. But this is only about the existence of the limit. If you replace $\lim$ by $\limsup$ then both are equivalent. 
3) Both the ratio and the ratio test are equivalent to using comparison with the series $\sum_n q^n$ for some $q$. So comparing with geometric series is as good as ratio and root test.
4) To use the comparison test it is good to have a good collection of series that we know about. Geometric $\sum_n q^n$, Harmonic, $\sum_n \frac{1}{n^r}$, and maybe some slower convergent/divergent series $\sum_n \frac{1}{n^r\ln(n)}$, $\sum_n \frac{1}{n^r\ln(n)\ln\ln(n)}$, ...
5) Finally, in the ratio and root test one must compute a limit. Since the two methods are essentially equivalent one chooses according to what resulting limit is more comfortable to compute.
In your example, there are mainly multiplications in the term, and constant exponents. Let us use ratio test (just because the limit is probably simple to compute and because maybe we don't know with what to compare. After all, applying ratio test is kind of mechanic). 
We compute $\lim \frac{\frac{n+1}{(2(n+1)+1)(2(n+1)-1)}}{\frac{n}{(2n+1)(2n-1)}}=1$.
Therefore the ratio test is inconclusive. There is no need to bother with the root test, as we know it will give the same answer. There is no need to compare with a geometric series as we know it will also give the same answer. So, checking the ratio test might not be a waste of time after all since it is telling us a few other techniques that will not work. So we go to compare with slowed converging/diverging series. the next in the collection are $\sum_n \frac{1}{n^r}$. Compare with them, maybe use the limit form of the comparison test. Enforce that the limit exists and is not zero. This will tell you that the right value for $r$ is $1$. So comparing with $\sum \frac{1}{n}$ gives us the answer. We should get that $\sum \frac{1}{n}$ and our series both converge or diverge. Since $\sum\frac{1}{n}$ diverges so does our series.
So, although in general it is not possible to tell an algorithm, for text-book examples and exam questions the methodology is rather mechanic. Most books and teachers don't point it out, thought. I think it is because they think exposing the mechanics of the process may subtract from the self-proclaimed beauty of mathematics. But the truth is that successful students are those that manage to grasp the mechanics of it. 

The ultimate goal in mathematics is to eliminate any need for intelligent thought.
Alfred N. Whitehead

