# $\{x_n\}$ is an unbounded increasing sequence. Prove that $\exists \lim_n {K_n\over n} =L \iff \exists \lim_n{n\over x_n} =L$, where $K_n$…

Let $$\{x_n\}$$ be an unbounded increasing sequence such that $$x_n \ne 0$$ and $$n\in\Bbb N$$. Let $$K_n$$ define the number of terms in $$\{x_n\}$$ such that: $$x_n \le n, n\in\Bbb N$$ Prove that if: $$\exists \lim_{n\to\infty} {K_n\over n} = L_1$$ then: $$\exists \lim_{n\to\infty} {n\over x_n} = L_2$$ and vice versa. And: $$L_1 = L_2$$

I've started with putting down given facts. First $$x_n$$ in increasing and unbounded: $$\forall M\in\Bbb R\ \exists n \in\Bbb N: x_n > M\\ x_{n+1} > x_n$$

This implies: $$\lim_{n\to\infty}x_n = +\infty$$

Next, we have $$K_n$$ which define a number of terms in $$x_n$$ which are less or equal than $$n$$. This means $$K_n$$ is never larger than $$n$$, which in terms imply: $$\exists \lim_{n\to\infty}{K_n\over n} = L \in [0, 1]$$

In case the limit exists then it must be somewhere in $$[0, 1]$$. Going back to the problem, what we want to prove is: $$\exists \lim_{n\to\infty}{K_n\over n} = L \iff \exists \lim_{n\to\infty}{n\over x_n} = L$$ And the problem splits into two parts, proving $$\implies$$ and proving $$\impliedby$$. This is where I'm not sure how to start.

To get some insight I decided to consider a couple of examples for $$x_n$$. Consider the following sequence: $$x_n = n - {1\over 2} = \left\{{1\over 2}, {3\over 2}, {5\over 2}, \dots \right\}$$

Clearly, the number of terms not exceeding $$n$$ is itself $$n$$, because every term of $$x_n$$ is less than $$n$$. So: $$K_n = \{1, 2, 3, \dots n\}$$

And then both limits exist: $$\lim_{n\to\infty} {K_n\over n} = 1 \\ \lim_{n\to\infty} {n\over n - {1\over 2}} = 1$$

Let's also try to consider the following sequence: $$x_n = n + {1\over 2} = \left\{{3\over 2}, {5\over 2}, \dots \right\}$$ Thus: $$K_n = \{0, 1, 2, \dots \} = n - 1$$ Therefore: $$\lim_{n\to\infty}{K_n\over n} = \lim_{n\to\infty}{n-1\over n} = 1$$ And: $$\lim_{n\to\infty}{n\over x_n} = \lim_{n\to\infty}{n\over n + {1\over 2}} = 1$$

The staments holds for both examples, but:

How do I prove this in general?

• @MartinR I've just looked in that question, but unfortunately the answer is not full in there – roman Mar 25 at 15:43

Choose $$j_0$$ sufficiently large such that $$x_j > 0$$ for $$j \ge j_0$$. For each $$j \ge j_0$$ there is a non-negative integer $$m_j$$ such that $$m_j < x_j \le m_j + 1 \, .$$ Then $$K_{m_j} < j \le K_{m_j+1}$$ and therefore $$\frac{K_{m_j}}{m_j+1} \le \frac{K_{m_j}}{x_j} < \frac{j}{x_j} \le \frac{K_{m_j+1}}{x_j} < \frac{K_{m_j+1}}{m_j}$$ $$(x_j)$$ is unbounded, so that $$j \to \infty$$ implies $$m_j \to \infty$$, and therefore $$\lim_{j \to \infty } \frac{K_{m_j}}{m_j+1} = \lim_{j \to \infty } \frac{K_{m_j+1}}{m_j} = L_1$$ and the conclusion follows with the squeeze theorem.
• Thank you for the answer, should it be $j \le K_{m(j) + 1}$ instead of $j \le K_{m(j+1)}$? – roman Mar 25 at 16:16
• Could you please elaborate on where it follows from that: $K_{m_j} < j < K_{m_j + 1}$. That's the only part left I'm trying to wrap my mind on. – roman Mar 25 at 16:47
• @roman: If $x_j \le m_j + 1$ then at least $j$ sequence elements ($x_1, x_2, \ldots, x_j$) are counted in $K_{m_j+1}$, therefore $K_{m_j+1} \ge j$. And if $x_j > m_j$ then there are at most $(j-1)$ sequence elements ($x_1, \ldots, x_{j-1}$) counted in $K_{m_j}$, so that $K_{m_j} < j$. – Martin R Mar 25 at 17:04