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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?

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

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    $\begingroup$ Thank you for the answer, should it be $j \le K_{m(j) + 1}$ instead of $j \le K_{m(j+1)}$? $\endgroup$ – roman Mar 25 at 16:16
  • $\begingroup$ @roman: Yes, indeed, that was a typo. $\endgroup$ – Martin R Mar 25 at 16:20
  • $\begingroup$ 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. $\endgroup$ – roman Mar 25 at 16:47
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    $\begingroup$ @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 $K_{m_j} < j$. $\endgroup$ – Martin R Mar 25 at 17:04
  • $\begingroup$ That's all clear now. Many thanks to you! $\endgroup$ – roman Mar 25 at 17:12

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