Let a sequence $\{l_n\}, n \in \Bbb N$ denote the amount of natural numbers $p \in \Bbb N$ which satisfy the following inequality: $$ 100n + 1 \le p^2 \le 100(n+1) $$ Prove $\{l_n\}$ diverges.
This is a bit tricky at first glance but lets try to track down what the inequality actually means. This is nothing but a set of intervals in the form $[100n + 1; 100(n+1)]$, or for the first several numbers: $$ [101, 200], [201, 300], [301, 400], \dots [100n + 1, 100(n+1)] $$
None of the first $10$ natural numbers get into the range $[101, +\infty]$. Let $N(s)$ be a function which takes a set and outputs the amount of elements in that set. Lets write down several terms of $\{l_n\}$: $$ l_1 = N(\{p: 101 \le p^2 \le 200\}) = N(\{11, 12, 13, 14\}) = 4 \\ l_2 = N(\{p: 201 \le p^2 \le 300\}) = N(\{15, 16, 17\}) = 3 \\ \dots \\ l_6 = N(\{p: 601 \le p^2 \le 700\}) = N(\{25, 26\}) = 2 \\ \dots $$
Clearly the number of squares which gets into the range with length $100$ is starting to decrease while $p$ is increasing, so eventually we will either get $1$ or $0$ for $l_n$ given $n$ is large enough.
This means we could choose two subsequences say $l_{n_k}$ and $l_{n_p}$ such that: $$ \lim_{k\to\infty} l_{n_k} = 1\\ \lim_{p\to\infty} l_{n_p} = 0 $$
Which would mean that the sequence $\{l_n\}$ is divergent.
The thoughts above are very informal and in my opinion are not rigorous enough to be considered a valid proof.
Could someone please show how those thoughts may have been formalized? I would also like to use a similar approach for $n^k + 1\le p^{k+1} \le (n+1)^k$ given some fixed $k \in \Bbb N$.
The strategy looks good, you just need to find those $2$ subsequences:
Here is a subsequence $\left\{l_{k^2+2k}\right\}_{k\in\mathbb{N}}$ such that $$\color{blue}{\left(10\cdot\left(k+1\right)-1\right)^2}= 100k^2+180k+81< \\ \color{red}{100\cdot \left(k^2+2k\right) + 1 } \leq p=\color{blue}{10^2\cdot(k+1)^2}= \color{red}{100\cdot\left(\left(k^2+2k\right)+1\right)}$$ where each $l_{k^2+2k}=1$, (in red are the boundaries, in blue $2$ consecutive perfect squares).
And a subsequence $\left\{l_{k^2+2k+1}\right\}_{k\in\mathbb{N}}$ such that $$\color{blue}{10^2\cdot\left(k+1\right)^2}< \\ \color{red}{100\cdot \left(k^2+2k+1\right) + 1} < \color{red}{100\cdot\left(\left(k^2+2k+1\right)+1\right)}=\\ 100\cdot\left(k^2+2k+2\right)<100k^2+220k+121= \color{blue}{\left(10\cdot(k+1)+1\right)^2}$$ where each $l_{k^2+2k+1}=0$ (for $k\geq 4$).