if $ 0 < y < 1$ show $y^n \rightarrow 0$ when $n \rightarrow \infty$ & if $ y > 1 $ then $y^{\frac{1}{n}} \rightarrow 1$

Question

i) if $ 0 < y < 1$ show $y^n \rightarrow 0$ when $n \rightarrow \infty$

ii) if $ y > 1 $ then $y^{\frac{1}{n}} \rightarrow 1$

I know it is not practice to ask two questions at once, but I figured since they are of similar ilk then it would be more helpful to combine them.

Attempt i)

I am already aware of one form of establishing this convergence by using the Monotone Convergence theorem. Our professor suggested another way to establish the convergence and I was trying to do it in that fashion.

Proof:

By the Archimedian property of the real numbers we know there exists a $p \in \mathbb{N}$ such that $0 < y < \frac{p}{p+1} < 1$. So it would be enough to show that: $$\Bigg(\frac{p}{p+1}\Bigg)^n \rightarrow 0$$

and then conclude by the squeeze theorem: $$y^n \rightarrow 0$$

I am trying to prove this using the definition of a limit of a sequence, so after expanding using the binomial theorem what I am trying to show is: $$\forall \epsilon >0, \ \exists \ N \in \mathbb{N} \ s.t.\ n \geq N \Rightarrow \Bigg| \frac{p^n}{p^n +np^{n-1} + \ldots + 1} - 0 \Bigg| < \epsilon$$

Notice: $p^n +np^{n-1} + \ldots + 1 > p^n + 1$..........

This is where I get stuck, I am trying to produce a N to satisfy my definition. I thought of possibly using logarithms but that was not fruitful. I am probably missing some simple trick, but alas I have not figured it out. What am I missing here?

Attempt ii)

For this question I want to use the Monotone convergence theorem to establish the result. The difficulty I am having is actually showing that $$y^{\frac{1}{n}} > y^{\frac{1}{n+1}}$$

Apparently by the ordered field axioms this already stands, but I don't recall establishing this result. If that is the case then It would be monotonically decreasing. But I don't think we've established this and I want to know how. What am I missing here?

Thank you for the help.

Answer 1

For i),

$$\Bigg| \frac{p^n}{p^n +np^{n-1} + \ldots + 1} - 0 \Bigg| < \Bigg| \frac{p^n}{p^n +np^{n-1}} \Bigg| = \Bigg| \frac{1}{1 +\frac{n}{p}} \Bigg|$$

For ii), since $y>0$, if not then $${(y^{\frac{1}{n}})}^{n(n+1)} \leq {(y^{\frac{1}{n+1}})}^{n(n+1)}$$ $$y^{n+1} \leq y^n$$ which is clearly impossible with $y>1$. Is this sufficient?

As an alternative way to do ii), write $y^{\frac{1}{n}}=1+\delta_n$. It suffices to show that $\delta_n \rightarrow 0$. By Bernoulli's inequality, $y=(1+\delta_n)^n \geq 1+n\delta_n$. For here $y$ serves as an upper bound the proof is finished.

This should be one of the standard ways for such limit.

Answer 2

Should this be your third attempt? We have $\ln(y)<0$ for $0<y<1$ and $\ln(y)>0$ for $y>1$. So for the first case

$$y^n=\exp(n\ln(y))\xrightarrow{n\to \infty}{0}\:\text{since}\; n\ln(y)\to-\infty$$ The second case is quite similar and I leave it for you.

Answer 3

For part $(a)$, we have that $$\left|\frac{p^n}{(p+1)^n}\right|\lt\left|\frac{p^n}{p^n+np^{n-1}}\right|=\left|\frac1{1+\frac np}\right|$$

Given $\epsilon\gt 0$, take $N\gt p\left(\frac 1\epsilon-1\right)$, and we're done.

Answer 4

This problem can be solved by elementary method without invoking transcendental functions like $\ln$ or $\exp$. I try part (i): Let $y\in(0,1)$. Define $a=\frac{1}{y}-1>0$. Then $y=\frac{1}{1+a}$. Observe that $(1+a)^{n}\geq1+na$, so $0\leq y^{n}\leq\frac{1}{1+na}\rightarrow0$ as $n\rightarrow\infty$. By squeeze theorem, it follows that $\lim_{n\rightarrow\infty}y^{n}=0$.

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For part (ii). Let $y>1.$ For each $n$, let $x_{n}=y^{\frac{1}{n}}-1>0$. We have $y=(1+x_{n})^{n}\geq1+nx_{n}\geq nx_{n}$. It follows that $0\leq x_{n}\leq\frac{y}{n}$. Therefore $\lim_{n\rightarrow\infty}x_{n}=0$ and hence $\lim_{n\rightarrow\infty}y^{\frac{1}{n}}=1$.