Prove n^(1/n) tends to 1 as n tends to infinity

[heshbon]

Homework Statement

Need to prove n^(1/n) tend to 1 as n tends to infinty

Homework Equations

The Attempt at a Solution

Have tried comparing to n^(1/n)=(1+h) and using binomial series but no joy..please help

 

[HallsofIvy]

Homework Statement

Need to prove n^(1/n) tend to 1 as n tends to infinty

Homework Equations

The Attempt at a Solution

Have tried comparing to n^(1/n)=(1+h) and using binomial series but no joy..please help

The sequence $n^{1/n}$, as n goes to infinity, converges to a if the function $x^{1/x}$ converges to a as x goes to infinity. If we set $y= x^{1/x}$ then ln(y)= (ln x)/x which is of the "infinity/infinity" form so we can use L'Hopital's rule.

 

[Pere Callahan]

One direct method that comes to my mind is to show that for any $\varepsilon>0$ there exist N such that
$$n^{1/n}\leq 1+\varepsilon$$
(It is easy to see that
$$n^{1/n}\geq 1$$
)
for all n>N.
The first equation is equivalent to $n\leq(1+\varepsilon)^n=1+n\varepsilon+\dots$

Do you see how to choose N?

 

[heshbon]

wow..briliantly simple using l'hopital...though i have not yet come across this theorem at uni...still will impress the tutors. thanks

 

[heshbon]

I can't see how to choose N...could you give me another hint?

 

[Pere Callahan]

The first equation is equivalent to $n\leq(1+\varepsilon)^n=1+n\varepsilon+\dots$
?

I should have included the next term in the binomial expansion

$$n\leq 1+n\varepsilon+\frac{n(n-1)}{2}\varepsilon^2\Leftrightarrow \dots$$
You just have to solve this for n>... and take the next larger integer for N.