What are the conventions for choosing the second series while Comparison Testing

Question

Given $\sum\limits_{n=1}^{\infty} \frac {\sqrt{n+2}} {2n^2+n+1} $

The author chose $b_n$ to be $\frac{1}{n^{3/2}}$

How did she choose that?

More generally, what are the standard practices when it comes to choosing $b_n$ ?

Answer 1

To prove a series $\sum a_n$ converges by comparison to another convergent series $\sum b_n,$ you want every term of the second dominating the absolute value of the corresponding one in the first, from some point on. That is, you want $|a_n|\le b_n$ for all $n>N,$ where $N$ is some positive integer. Then you will have that $\sum |a_n|\le \sum b_n=\beta\in\mathrm R,$ so that you have a monotonically increasing sequence bounded above by $\beta,$ and hence convergence to some value $\le\beta$ follows.

Similarly, to use the comparison test to prove divergence, you want to show that every term of a divergent series $\sum c_n$ is less than the absolute values of your series $\sum a_n$ from some point on. Then you will have that $\sum|a_n|\ge\sum c_n=\pm\infty. $ It follows that your series itself becomes arbitrarily large.

How do we know which series to compare to? That comes from experience. As you learn about more and more series and file away their convergence or divergence in your mind, when you see similar series you will know how to compare. It's just like integrating -- mastering it comes only from experience. No exact algorithm as you seek exists.

In relation to the example you gave, note that the numerator, when $n$ becomes very large, is approximately like $\sqrt n=n^{1/2}$ (that is, the $1$ may be neglected). Also, the denominator behaves like $n^2$ as $n\to+\infty.$ Thus the fraction is asymptotically equivalent to $n^{1/2}/n^2=n^{-3/2},$ or $$\frac{1}{n^{3/2}},$$ whose convergence may be easily deduced.

Answer 2

There is a very useful series to keep in mind when considering convegrence of series. That series is $\sum\frac{1}{n^p}$, since it is known to converge when $p>1$ and diverge when $p\leq 1$.

Using that result if you can compare $0\leq a_n\leq \frac{1}{n^p}$ when $p>1$, then $\sum a_n$ converges. On the other hand if $a_n\geq \frac{1}{n^p}$ while $p\leq 1$, then $\sum a_n$ diverges.

A more natural method to decide in this case with what to compare, is the limit comparison test for non-negative series. Recall that if $\frac{a_n}{b_n}\rightarrow L\in(0,\infty)$, then $\sum a_n$ converges if and only if $\sum b_n$ converges.

Notice that in your case

$$ \frac{ \frac{\sqrt{n+2}}{2n^2+n+1} }{\frac{1}{n^{3/2}}}\rightarrow \frac{1}{2}. $$

In general saying $a_n\sim \frac{1}{n^p}$ solves the question of convergence of a non-negative series.