If $lim\: u_n=0$ then show that $\forall \: \epsilon \:\exists N\: n\geq N \Rightarrow (n-N)u_n \leq \epsilon$

Question

Suppose $u_n$ is positive and strictly decreasing. Then show that if $\lim u_n=0$ that $$\forall \: \epsilon \:\exists N\: n\geq N \Rightarrow (n-N)u_n \leq \epsilon$$

This is my attempt. Let $N \in \mathbb{N}$, let $\epsilon \geq 0$ and let's define a sequence $v_n$ such that $v_n=\frac{\epsilon}{n-N}$. The limit of this sequence is zero as $n$ tends to infinity.

Thus $\lim u_n v_n = 0$, and so we have $(n-N)u_nv_n=\epsilon u_n$, as $\epsilon$ is fixed, its limit is still 0. Thus we can conclude that the limit of $(n-N)u_nv_n$ is $0$. Thus the limit of $(n-N)u_n$ is zero. And so $\forall \epsilon \: \exists N \: \forall n\geq N \: lim\:(n-N)u_n=0$.

I just have a bad feeling about my attempt. I don't think I can state that if a limit of a product is $0$ then the limits of both terms is $0$. Any criticism would be highly appreciated.

Answer 1

Let $N \in \mathbb{N}$, let $\epsilon \geq 0$ and let's define a sequence $v_n$ such that $v_n=\frac{\epsilon}{n-N}$. The limit of this sequence is zero as $n$ tends to infinity.

This limit is indeed zero, alas you did not proof it by an epsilon-delta style proof.

Thus $\lim u_n v_n = 0$,

Yes, as the individual limits are supposed to exist and both are zero, the limit of the product exists and is zero: $$ 0 = 0 \cdot 0 = \lim u_n \lim v_n = \lim u_n v_n $$

and so we have $(n-N)u_nv_n=\epsilon u_n$, as $\epsilon$ is fixed, its limit is still 0.

Indeed if $u_n$ has limit zero and $\epsilon$ is a fixed number then $\epsilon u_n$ has limit zero as well: $$ \lim \epsilon u_n = \epsilon \lim u_n = \epsilon \cdot 0 = 0 $$

Thus we can conclude that the limit of $(n-N)u_n v_n$ is $0$.

Yes, by noting the equality $$ (n-N) v_n u_n = (n-N) \frac{\epsilon}{n-N} u_n = \epsilon u_n $$

Thus the limit of $(n-N)u_n$ is zero.

Your argument seems to be $$ 0 = \lim \, ((n-N) u_n) v_n \Rightarrow \lim (n-N) u_n = 0 \quad (*) $$

Suppose $u_n$ was defined as $$ u_n = \frac{1}{n-N} $$ then this $u_n$ is positive and is strictly decreasing and has $\lim u_n = 0$, thus this $u_n$ is meeting the constraints of the problem and we would have $$ \lim ((n-N) u_n)v_n = \lim \left((n-N) \frac{1}{n-N} \frac{\epsilon}{n-N} \right) = \lim \frac{\epsilon}{n-N} = 0 $$ but $$ \lim (n-N) u_n = \lim (n-N) \frac{1}{n-N} = \lim 1 = 1 \ne 0 $$ so your argument $(*)$ is not valid for this $u_n$.

Answer 2

You're right to have the bad feeling, in that $(n-N)u_n v_n \to 0$ does not in general imply $(n-N) u_n \to 0$. (Let $v_n = \frac{1}{n^2}$, and $u_n = \frac{1}{\sqrt{n}}$.) Whether that actually holds in this specific circumstance, with $v_n = \frac{\epsilon}{n-N}$, is what you're trying to prove.

Your aim is to show that if $(u_n)$ is positive and strictly decreasing, then $(x_n) \to 0$ where $x_n = (n-N) u_n$. Does that look plausible to you?

Hint: some of the most useful sanity checks in undergrad analysis are the function $x \mapsto e^x$ and the sequences $(n)$, $(\frac{1}{n})$, $(\frac{1}{n^2})$, and $(\frac{1}{\sqrt{n}})$.