A sequence whose sum is infinite but whose sum of squares is not? I am thinking of positive sequences whose sum is infinite but whose sum of squares is not?
One representative sequence is $$x[n] = \frac{a}{n+b},$$ where $a$ and $b$ are given real numbers such that $a>0$ and $b\ge0$.
I know that there will be infinitely many more sequences $x[n]$ such that $x[n]\ge0, ~x=1, 2, ...$, $\sum x[n] = \infty$, and $\sum (x[n])^2 <= M$ for a sufficiently large constant value $M$.
Can you give me some examples? If possible, I would really appreciate it if you could tell me how to find these sequences (i.e., methodology of how to find).
 A: Essentially copying off wikipedia, the property you are asking for is related to something called the $\ell^p$ sequence space. Specifically, for some base field, say the reals, for $0 < p < \infty$, the $\ell^p$ space consists of all sequences $(x_n)$ satisfying $\sum_n |x_n|^p < \infty$. You are looking for real sequences that lie in the $\ell^2(\mathbb{R})$ space but not in $\ell^1(\mathbb{R})$ space. As far as I know, if $\mathbb{R}$ is replaced by a closed interval $[a, b]$, then we have $\ell^1([a, b]) \subseteq \ell^2([a, b])$, see here. There are various related questions on mathSE as well. Hope this helps your googling journey.
A: Let's first notice, that if
$$\limsup_{n\to\infty} \frac{x_{n+1}}{x_n} < 1$$
then both series $\sum_{n=1}^\infty x_n$ and $\sum_{n=1}^\infty x_n^2$ converge. On the other hand, if
$$\liminf_{n\to\infty} \frac{x_{n+1}}{x_n} > 1$$
then both series $\sum_{n=1}^\infty x_n$ and $\sum_{n=1}^\infty x_n^2$ diverge. Therefore you need as a necessary (but not sufficient) condition
$$ \liminf_{n\to\infty} \frac{x_{n+1}}{x_n} \le 1 \le \limsup_{n\to\infty} \frac{x_{n+1}}{x_n}.$$
To find the convergence of such series, you can often use the Raabe's test. We define
$$ y_n = n\left(\frac{x_n}{x_{n+1}} -1\right)$$
$$ z_n = n\left(\frac{x^2_n}{x^2_{n+1}} -1\right) $$
The series $\sum_{n=1}^\infty x_n$ diverges while $\sum_{n=1}^\infty x_n^2$ converges if
$$ \limsup_{n\to\infty} y_n \le 1, \qquad \liminf_{n\to\infty} z_n > 1 $$
For example, any sequence which has the asymptotic behavior $x_n \sim n^{-\alpha}$, $\alpha \in(\frac12, 1]$ will give you
$$\lim_{n\to\infty} y_n = \alpha \le 1 $$
$$\lim_{n\to\infty} z_n = 2\alpha > 1 $$
so it will satisfy these conditions.
A: As the answer from Gareth Ma points out, look for sequences in $\ell^2 \setminus \ell^1$. This will be your set of sequences.
Concretely speaking, take $\{a_n : a_n \geq 0\}$ such that $\sum_n a_n < \infty$. Now look at $\sum_n \sqrt{a_n}$ to generate the sequences you want. This method may sometimes work. Using this method we have $a_n = \frac{1}{n^{1+\epsilon}}$ which works. We know that $\sum_n \frac{1}{n^{1+\epsilon}} < \infty$ for every $\epsilon>0$. This can be seen from the fact that $\sum_{n \geq k+1} \frac{1}{n^{1+\epsilon}} \leq \int_k^{\infty} \frac{dx}{x^{1+\epsilon}} < \infty$ for $k \geq 1$. Now $\sum_n \frac{1}{n^{\frac{1+\epsilon}{2}}} = \infty$ for every $0<\epsilon \leq 2$.
Specifically look for $a_n$ such that $$\lim_{n \rightarrow \infty} \frac{|a_{n+1}|}{|a_n|} = 1$$ or else it wont work.
This is because $$\lim_{n \rightarrow \infty} \frac{|a_{n+1}|}{|a_n|} > 1 \iff \lim_{n \rightarrow \infty} \frac{\sqrt{|a_{n+1}|}}{\sqrt{|a_n|}} > 1 $$. Hence both $\sum_n a_n = \infty $ and $\sum_n \sqrt{a_n} = \infty$ and because $$\lim_{n \rightarrow \infty} \frac{|a_{n+1}|}{|a_n|} < 1 \iff \lim_{n \rightarrow \infty} \frac{\sqrt{|a_{n+1}|}}{\sqrt{|a_n|}} < 1 $$. Hence both $\sum_n a_n < \infty $ and $\sum_n \sqrt{a_n} < \infty$.
A Method to Generate more such sequences:
Let $a_n \geq 0$ and is monotonic and bounded and if $\{b_n\}$ is such that $b_n \geq 0$ and $\sum_n b_n < \infty$.
Further if $\sum_n a_n < \infty$ and $\sum_n \sqrt{a_n} = \infty$ then $s_n = \sqrt{a_n} + b_n$ works since $\sum_n s_n = \infty$ and $\sum_n s_n^2 < \infty$ by Abel's convergence test. So you can practically generate infinite number of such sequences using a single such $\{a_n\}$ sequence by adding any positive convergence sequence $\{b_n\}$. As an example $a_n = \frac{1}{n^{1+\epsilon}}$ works in this context.
A: The standard example is the sequence $x_n = 1/n$.
We have $\sum_{n=1}^N \frac{1}{n} \simeq \log N$ and so $\sum_{n=1}^N \frac{1}{n}$ diverges.
At the same time $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$ and the squared series converges.
More generally we have $\sum_{n=1}^N \frac{1}{n^p} \simeq p N^{p-1}$ for $0<p <1$ and so the series diverges. While for $p >1$ the series converges. You can use these two facts to come up with other examples.
