Suppose $\{x_n\}$ is bounded. Prove $\frac{x_n}{n^k} \rightarrow 0$, as $n \rightarrow \infty$, for all $k \in \mathbb{N}$.

Question

I've came up with the following proof for the question in the title. I'm looking for suggestions on presentation and on methodology as I'm relatively new to analysis. In particular, my book says a sequence converges to $a$ if and only if for any $\epsilon > 0$ there's a $N \in \mathbb{N}$ such that $n \geq N$ implies $|x_n - a| < \epsilon$. In my proof I weaken this a bit by only taking $n > N$. Is this sort of deviation valid?

Let $\epsilon > 0$. Suppose $\{x_n\}$ is bounded. Then there exists some $C > 0$ with $C \in \mathbb{R}$, where $|x_n| \leq C$ for all $n \in \mathbb{N}$. Set $N = \frac{C}{\epsilon}^{\frac{1}{k}}$ and suppose that $n > N$. It follows that $$|\frac{x_n}{n^k}| = \frac{|x_n|}{n^k} \leq \frac{C}{n^k} < \frac{C}{(({\frac{C}{\epsilon})^{\frac{1}{k}}})^k} = \epsilon$$ for all $n > N$.

Answer 1

Take $M\in \Bbb R^+$ such that $\forall n\in \Bbb N\,(M>|x_n|).$

For $n,k\in \Bbb Z^+$ we have $|\frac {x_n}{n^k}|=$ $\frac {|x_n|}{|n^k|}=$ $\frac {|x_n|}{|n|^k}=$ $ \frac {|x_n|}{n^k}\le\frac {|x_n|}{n}<$ $\frac {M}{n}.$

A direct consequence of the $definition$ of $\Bbb R$ is that for any $x\in \Bbb R$ there exists $m\in \Bbb Z^+$ such that $x<m.$

So, given $\epsilon>0 ,$ take $ m_{\epsilon}\in \Bbb Z^+$ such that $m_{\epsilon}>\frac {M}{\epsilon}.$ Then $$ \forall n\in \Bbb N\,\left(n\ge m_{\epsilon}\implies\left|\frac {x_n}{n^k}\right|<\frac {M}{n}\le \frac {M}{m_{\epsilon}}<\epsilon\right).$$

Answer 2

In the definition of convergence, there are two modifications to know about that can be made without consequence and often reduce the cognitive load of the definition of limits:

1. One of them, as you noted, is that you can change $\forall n\ge N$ to $\forall n> N$.
2. The other is that instead of requiring the strict inequality $|a_n - L| < \epsilon$, mere inequality suffices $|a_n - L| \le \epsilon$.

Clearly in 1., the weak inequality version implies the strong inequality version. The strong inequality version for 1. also implies the weak inequality version by taking $N_\text{weak} = N_\text{strong}+1$.

In 2., the strong clearly implies the weak. The weak inequality in 2. implies the strong by taking $\epsilon_\text{strong} = \epsilon_\text{weak}/2$.

Answer 3

If we take $0 \in \mathbb{N}$, then the statement isn't true for all $k \in \mathbb{N}$. It does not work for $k = 0$. For example, the sequence $(1, 1, 1, \cdots)$ is bounded, but doesn't converge to $0$. It is true for all positive integers $k$, however. So let $k$ be a positive integer.

We need to be careful to define $N$ such that it is indeed a natural number. With the definition $N = \left(\frac{C}{\epsilon}\right)^{\frac{1}{k}}$, this is not necessarily the case.

To fix this, we need to find a natural number $N$ which is greater than or equal to $\left(\frac{C}{\epsilon}\right)^{\frac{1}{k}}$, so that we can slightly alter the steps given in the question to get the inequality we desire. A candidate is $ N = \bigg\lceil\left(\frac{C}{\epsilon}\right)^{\frac{1}{k}}\bigg\rceil$. We note that $N \neq 0$, since $C > 0$.

We have $N \geq \left(\frac{C}{\epsilon}\right)^{\frac{1}{k}}$, so $N^k \geq \frac{C}{\epsilon}.$

Following the steps given in the question, for each $n > N$, we have

$$\bigg|\frac{x_n}{n^k}\bigg| = \frac{|x_n|}{n^k} \leq \frac{C}{n^k} < \frac{C}{N^k} \leq C \cdot \frac{\epsilon}{C} = \epsilon.$$ Therefore, $$\bigg|\frac{x_n}{n^k}\bigg| < \epsilon.$$

This is the inequality that we desire, which is true for any $n > N$, where $N \in \mathbb{N}$.