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

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

• Key point to notice is that $x>0$ and $x^{-1/5}<\epsilon$ is equivalent to $x^{1/5}>\frac{1}{\epsilon}$, which in turn is equivalent to$x>\frac{1}{\epsilon^5}$. – Nick Peterson Mar 6 '17 at 21:10

We can write $n+1$ instead of $n$ so with $0<\varepsilon<1$ $$0\leq(1+n)^{-\frac15}=\dfrac{1}{(1+n)^{\frac15}}\leq\dfrac{1}{1+\frac15n}=\dfrac{5}{5+n}\leq\varepsilon$$
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 $1/(\epsilon^5)$.
• No. You cannot take $N=\epsilon$ because $N$ is a natural number. – Andres Mejia Mar 6 '17 at 21:21