Proving $\lim \limits_{n \to \infty} n^{-\frac{1}{5}} = 0$ formally using $\delta - \epsilon$

Question

Prove $\lim \limits_{n \to \infty} n^{-\frac{1}{5}} = 0$ formally using $\delta - \epsilon$

It's been a while since I've formally proved limits (4-5 months?) and it was never a strong point of mine.

I understand I have to pick an N s.t. $n\gt N$, and of course $\left| n^{-\frac{1}{5}}\right| \lt \epsilon$

So my N has to satisfy both of those inequalities.

I was thinking of taking $N= \epsilon $,

therefore:

$\left|n^{-\frac{1}{5}}\right |\lt \left|N^{-\frac{1}{5}}\right| = \left|\epsilon^{-\frac{1}{5}}\right | \lt \epsilon $

I feel like the last part might be incorrect, since taking the fifth root will make the denominator smaller, and the number larger.

I'd appreciate any input, if it's wrong as to why it's wrong, and if it's right, how I could feel more comfortable with the last inequality

Answer 1

Use the Archimedean property:

let $\epsilon>0$. When is $1/(n^{1/5})<\epsilon$? well, when $n>1/\epsilon^{5}$. The existence of such an $N \in \mathbb N$ is ensured by the Archimedean property, and all $n>N$ are also greater than $1/(\epsilon^5)$.

Answer 2

We can write $n+1$ instead of $n$ so with $0<\varepsilon<1$ $$0\leq(1+n)^{-\frac15}=\dfrac{1}{(1+n)^{\frac15}}\leq\dfrac{1}{1+\frac15n}=\dfrac{5}{5+n}\leq\varepsilon$$