How do you test for convergence for $ \sum {\dfrac{{{{( - 1)}^{n - 1}}}}{{2n + 1}}} $ for $ n = $ $ 1 $ to infinity?

Hint: As we know that convergence in mathematics, is a property of approaching a limit when the variable in a function increases or decreases. It is exhibited by some of the infinite functions and series. In this question, we will break the parts in two and then solve it. We should know that if the larger series converges, then the smaller series also converges, and if the smaller series diverges, then the larger series diverges.

Complete step-by-step answer:
In the above question we can write
 $ \sum\limits_{n = 1}^\infty {\dfrac{{{{( - 1)}^{n - 1}}}}{{2n + 1}}} $ .
We will break the fraction into two parts without changing the original form: $ \sum\limits_{n = 1}^\infty {{{( - 1)}^{n - 1}} \times \dfrac{1}{{2n + 1}}} $ .
Let us assume $ \sum\limits_{n = 1}^\infty {\dfrac{1}{{2n + 1}} = {a_n}} $ , now we can write the given series in the form $ \sum\limits_{n = 1}^\infty {{{( - 1)}^{n - 1}}{a_n}} $ .
From the above we can see that $ {a_n} \geqslant 0 $ for all natural numbers $ n $ and is a sequence that decreases to zero. As we know that in alternating series, there are terms that decrease in absolute value as a convergent series.
Hence the alternating series test implies that $ \sum\limits_{n = 1}^\infty {\dfrac{{{{( - 1)}^{n - 1}}}}{{2n + 1}}} $ converges.

Note: We should know that the alternating series test can only tell us if the alternating series converges or not. It says nothing about the positive term series. It does not tell us whether a series is absolutely convergent or conditionally convergent. Before solving this type of question we should be fully aware of the convergent and divergent series.