The question is regarding how to solve a general set of series. If a series whose nth terms goes like this a= (-1)^n/ (n^3+2n) , how should I find whether the series is convergent or not.
Since the nth term is (-1)^n/(n^3+2n) , Can I compare it with a series b= 1/(n^3) and say that a is convergent since b is convergent. What I don't understand is this -
The p series test says that if limit (a/b) as n tends to infinity exists and is not 0 and if b is convergent then so is a. So can I say for the above example that limit (a/b) as n tends to infinity is either 1 or -1 ( both of which are finite and non zero) and since b is convergent so is a. I don't think this should be correct because the limit should be a single number. But I found a similar answer to an almost similar question here on Stack. (Final answer . I thought that for the answer to be ' convergent ' and the method to be the comparison test , the argument would be like this )
Or should I apply Leibniz's test. The nth terms as n tends to infinity tends to 0. But what about (n+1)th term <=nth term condition ? Wouldn't the above condition depend on n since we have (-1)^n sitting in the numerator and hence (n+1)th term can be positive or negative depending upon n and hence the condition (n+1)th term <=nth would vary . Am I wrong ? The last line is confusing me a lot.
I have written all the points which were causing major conceptual confusion in bold. Help regarding the solution laying stress on and explaining the points in bold is needed. Thanks in advance.
This is what is known as the Limit Comparison Test, (not the $p$-series test as you had written). This is true given a few extra conditions, namely $a_n\geq 0$ and $b_n\geq 0$. One can generalize it a bit further, as is done in this related question, but care needs to be taken. In particular things can break where one of the series is alternating in sign but not the other.
You suggested that by letting $a_n = \frac{(-1)^n}{n^3+2n}$ and $b_n=\frac{1}{n^3}$ and taking the limit of the ratio, "the limit is either $1$ or $-1$" but... as you allude to, the limit is only ever one number at a time! The sequence $(-1,1,-1,1,-1,1,-1,1,\dots)$ by definition diverges. In other words, the limit does not exist. Since the limit does not exist, the test above cannot successfully be directly applied. That doesn't mean that it tells us it diverges, it just means that we don't get any information about it directly.
(one can use the limit comparison test after deciding to discuss the absolute convergence of the series and there is no issue at that point)
The Leibniz Test (alternating series test) does actually apply here and give an immediate conclusive answer.
In your case, letting $a_n = \frac{1}{n^3+2n}$ we see that the series in question is exactly of the form $\sum\limits_{n=1}^\infty (-1)^na_n$. Since $a_{n}\frac{1}{n^3+2n}>\frac{1}{(n+1)^3+2(n+1)}=a_{n+1}$ and $\lim\limits_{n\to\infty}a_n=0$ the test applies.
You say "Wouldn't the above condition depend on n since we have (-1)^n sitting in the numerator..." The answer is that there is no $(-1)^n$ in the numerator of $a_n$ at all. The $(-1)^n$ is separate from $a_n$. All terms of $\frac{1}{n^3+2n}$ are positive and do not have a $(-1)^n$ involved anywhere.
While the Leibniz test us useful to tell us about conditional convergence for an alternating series, it is not powerful enough to tell us about absolute convergence. In many exercises in introductory calculus, you will not only be asked if a series converges, but will also be asked if it does so absolutely.
Absolute convergence implies convergence. Convergence however does not imply absolute convergence.
For your series in question, as alluded to in the comments above, $\sum\limits_{n=1}^\infty\left|\frac{(-1)^n}{n^3+2n}\right| = \sum\limits_{n=1}^\infty\frac{1}{n^3+2n}\leq \sum\limits_{n=1}^\infty\frac{1}{n^3}$ which converges by the $p$-series test. Thus, the original series is absolutely convergent by direct comparison and is therefore convergent as well.