How Can You Prove the Limit of (n-1/n) as n Approaches Infinity Equals 1?

[Ed Quanta]

How do you prove that the lim n->infinity (n-1/n)=1?

I know that the definition of convergence for a sequence xn is for all E>0, there exists an N is an element of the set of a natural numbers and there exists a n>=N, such that lxn-Ll<E.

Is it sufficient to just show that 1 is the least upper bound, and that

1- (n-1/n) <E and thus 1-E<(n-1/n) and then n-nE<n-1 and ultimately there is an n such that n-nE+1<n

 

[HallsofIvy]

I am tempted to say "you don't- it's not true!". But you mean (n+1)/n NOT
"n+(1/n)" which is how "n+ 1/n" would normally be interpreted.

(n+1)/n= 1+ 1/n . It should be easy for you to prove that 1/n-> 0 as n-> infinity.

Yes, you can do it as |1- (n-1)/n|= |(n-n+1)/n|= 1/n< E- that's much simpler than
multiplying by the denominator,n, before simplifying.

 

[M98Ranger]

and so there exists an N is an element of the set of a natural numbers and there exists a n>=N, such that l(n-1/n)-1l<E?

Yes, your approach seems correct. To prove that the limit of (n-1/n) as n approaches infinity is 1, we need to show that for any given epsilon (E), there exists a natural number N such that for all n>=N, the difference between (n-1/n) and 1 is less than E.

In your proof, you have shown that for any given E, there exists a natural number N (specifically, N = 1/E) such that for all n>=N, the difference between (n-1/n) and 1 is less than E. This is exactly what we need to show for convergence.

So, yes, your approach is sufficient to prove that the limit of (n-1/n) as n approaches infinity is 1.