Question

Epsilon Delta proof (shudders).
Show that n^1/n -> 1 as n-> ∞

Answer 1

Let \(\rm \large h_n = n^{1/n} -1 \)

Then\[\rm \large n = ( 1 + h_n ) ^n
\\ \large = 1 + nh_n + n(n-1) \frac{(h_n)^2}{2!} ... + (h_n)^n > 1 + nh_n \]
From which it follows \[ \rm \large n > 1 + nh_n
\\ \large \implies h_n < \frac{n-1}{n}
\\ \large \implies h_n < 1 - \frac 1 n < 1 + \frac 1 n ~~~~ \text{ since } n > 0
\] Then \[ \large \rm | h_n - 0 |= h_n < 1+ \frac 1 n < \epsilon~
\\ \large \rm \text{ if } n \ge N = \frac{1}{ \epsilon -1 }+1 \]
We can assume that \( \epsilon \neq 1 \) above

Answer 2

There will be a problem... since epsilon can be as small as we please

Answer 3

the inequality
1 - 1/n < e
did not produce
n > f(e)

Answer 4

1+1/n -> 1 and cannot be made close to zero

Answer 5

right...

Answer 6

So i can't salvage this proof?

Answer 7

Yeah, the method I used, thats correct.. elisaab told the correct thing, 1+1/n is always greater than 1

Answer 8

ok this is a bad proof since e can be less than 1
but 1 + 1/n is always greater than 1

Answer 9

ANy ways, you tried to give an alternative proof.. Its always good to try something new

Answer 10

1 - 1/n < e
1 - e < 1/n
n < 1 / (1 - e )

there might be a way to make n > something

Answer 11

Oh, in the proof we need \( \rm h_n\) to get as close to zero (the limit)
as possible?

That is not possible with \( \large \rm 1 - \frac 1 n \) or \( \large \rm 1 + \frac 1 n \)

Answer 12

Or in other words, we need to bound 1\( \pm \) 1\n
by epsilon, but epsilon goes to zero

Answer 13

right.

Answer 14

and since n goes to infinity, that can't go to zero

Answer 15

I like how in your proof you have | h(n) - 0 | < e
in general it seems like you can always do that

i.e. if lim f(n) = L , then h(n) = f(n) - L
and then find the limit of this.

This seems to simplify things because you can concentrate just on the epsilon