Determine the convergence of a series.

Question

Here is the series: $$ \sum_{n = 1}^{\infty}\frac{\sqrt{n + \sqrt{n + \sqrt{n}}}}{(n + (n + n^2)^2)^2}$$ The method I use to determine this series is comparison test which is that I construct the following sequence : $$ a_n = \frac{\sqrt{3n}}{n^8}$$ Which forms a convergent series where each term is greater than the terms in the series above so that I determine the series above is converge. However, I don't know whether if I am right or not. Therefore if I am wrong please tell me how to do it correctly or if I am correct please confirm with me or providing me an alternative method to determine the convergence of the series above for discussion. Thanks.

Answer 1

Honestly, unless there is an explicit instruction to use some test, I prefer to think about these kinds of series in terms of the limit comparison test (LCT), instead of the comparison test (CT).

The usual statement of the LCT is something like this: Suppose that $\{ a_n \}$ and $\{ b_n\}$ are sequences with $a_n \ge 0$, $b_n > 0$ for all $n$. If $\lim_{n\to +\infty} a_n/b_n$ exists and is nonzero, then $\sum a_n$ and $\sum b_n$ converge together, or diverge together.

The LCT cares less about the direction of the inequality (unlike with the CT where you have to verify certain inequalities which can be annoying), and more about the asymptotics, which makes it a lot more powerful. As for looking for the appropriate $b_n$ to use as a point of comparison? The usual idea is to look at the most dominant terms (i.e., the terms that blow up to infinity fastest) in the numerator and denominator.

In your example, the dominant term in the numerator is $\sqrt{n}$, while the dominant term in the denominator is $n^8$. This suggests that we use $b_n = \sqrt{n}/n^8 = n^{-15/2}$, which indeed works nicely here. We get $\lim_{n\to+\infty} a_n/b_n = 1$, and we know $\sum b_n$ converges by the $p$-test. Thus, so does the original series.

Answer 2

This method have own name Direct comparison test and states following:

If series $\sum b_n$ converges and $0 \leqslant a_n \leqslant b_n$ for sufficiently large $ N \in \mathbb{N}, n> N$, then $\sum a_n$ aslo converges.

Holds $\sum a_n \leqslant \sum b_n$ if comparison is $\forall n \in \mathbb{N}$.

If $\sum a_n$ diverges, then $\sum b_n$ is divergent.

In book: Murray H. Protter, Charles B. Jr. Morrey - Intermediate Calculus-Springer (2012)- page 105, Theorem 9.

Answer 3

Your solution is fine, but you feel a bit insecure, let me show why the test works: a series $\sum_{k= 1}^\infty a_k$, by definition, represent the limit of the sequence of it partial sums $\{s_n\}_{n\in \mathbb N}$, for $s_n:=\sum_{k=1}^na_k$.

When each $a_k$ is positive then the sequence $\{s_n\}_{n\in \mathbb N}$ is a sequence of positive real numbers that is strictly increasing and so it can be shown that it converges if and only if its bounded.

If $a_k:=\sqrt{k+\sqrt{k+\sqrt{k}}}/(k+(k+k^2)^2)^2$ then its easy to see that $0\leqslant a_k\leqslant k^{-2}$ for each $k\in \mathbb N $, and so

$$ 0\leqslant s_n\leqslant \sum_{k=1}^n k^{-2}\quad \text{ for each }n\in \mathbb N \\ \text{ and }\quad \sum_{k=1}^n k^{-2}\leqslant \sum_{k=1}^\infty k^{-2}=\frac{\pi ^2}{6}\quad \text{ for each }n\in \mathbb N \\ \text{ therefore }\quad 0\leqslant s_n\leqslant \frac{\pi ^2}{6}\quad \text{ for each }n\in \mathbb N $$

$\Box$