Proving lim[(2n+1)/(5n-2)] = 2/5 using the ϵ-N definition of limit

[Askhwhelp]

ϵ-N definition of limit

Use the ϵ-N definition of limit to prove that lim[(2n+1)/(5n-2)] = 2/5 as n goes to infinity.

The way I do it is Let ∊ > 0 be given. Notice N ∈ natural number (N) which satisfies {fill this box later}< N. It follows that if n>=N, then n > {fill this box later}, so for such n, |(2n+1)/(5n-2)-2/5| = |9/(25n-10)| = 9/5|1/(5n-2)|

I am supposed to get to a something that is less than ∊

How to make this to less than ∊?

 

[Dick]

Use the ϵ-N definition of limit to prove that lim[(2n+1)/(5n-2)] = 2/5 as n goes to infinity.

The way I do it is Let ∊ > 0 be given. Notice N ∈ natural number (N) which satisfies {fill this box later}< N. It follows that if n>=N, then n > {fill this box later}, so for such n, |(2n+1)/(5n-2)-2/5| = |9/(25n-10)| = 9/5|1/(5n-2)|

I am supposed to get to a something that is less than ∊

How to make this to less than ∊?

You are over halfway there. If you want 9/5|1/(5n-2)|<ε, you want 1/|5n-2|<5ε/9. 5n-2 is positive so you can drop the absolute value. Now turn it into an inequality for n.

 

[Askhwhelp]

Limit and Convergence

Suppose a_n > 0 and b_n > 0 for all n in natural number (N). Also, lim a_n/b_n = 0 as n goes to infinity. Then the sum of a_n converges if and only if the sum of b_n converges ...both from 1 to infinity.

My approach is that lim a_n/b_n = 0 means that there exists N in natural number (N) for which |a_n/b_n - 0| < 0 for all n >= N. Then 0 < a_n < 0. The sum of a_n from 1 to infinity is 0. So The sum of a_n from 1 to infinity is convergent.

Is this proof that easy or I miss something?

 

[haruspex]

Suppose a_n > 0 and b_n > 0 for all n in natural number (N). Also, lim a_n/b_n = 0 as n goes to infinity. Then the sum of a_n converges if and only if the sum of b_n converges ...both from 1 to infinity.

The 'only if' part is clearly false. Pick a_n = 2^-n, b_n = 1. Should it have said lim a_n/b_n = c > 0?