Prove √a(n) Converges to 0 | Stuck on Proof

[CoachBryan]

I've been messing with this proof for while and I'm stuck on this. I've started with a(n) converges to 0, let epsilon > 0, then there exists an n0 in N such that for all n >= n0.

I'm stuck here thus far. Any help? Thanks for your time.

 

[Mark44]

I've been messing with this proof for while and I'm stuck on this. I've started with a(n) converges to 0, let epsilon > 0, then there exists an n0 in N such that for all n >= n0.

I'm stuck here thus far. Any help? Thanks for your time.

The rest of your thought is
For all n >= n₀, ##\sqrt{a_n} < \epsilon##.

What are the given conditions? Is it a_n converges to 0? You have a_n = 0 in the title.

 

[CoachBryan]

The rest of your thought is
For all n >= n₀, ##\sqrt{a_n} < \epsilon##.

What are the given conditions? Is it a_n converges to 0? You have a_n = 0 in the title.

Yes it converges to zero

 

[deluks917]

Can you prove that sqrt(a_n) is eventually less than 1/100 ?

How about that sqrt(a_n) is eventually less than 1/(1000000) ?

 

[blue_raver22]

I understand the importance of being thorough and precise in our mathematical proofs. In this case, we are trying to prove that √a(n) converges to 0, which means that as n approaches infinity, the value of √a(n) gets closer and closer to 0.

To start, we can use the definition of convergence to help guide our proof. We know that for a sequence to converge to a limit L, we must have that for any small positive number ε, there exists a positive integer N such that for all n ≥ N, the absolute value of the difference between a(n) and L is less than ε.

In this case, our limit L is 0, so we can rewrite our definition as: for any small positive number ε, there exists a positive integer N such that for all n ≥ N, the absolute value of √a(n) is less than ε.

Now, let's use the fact that a(n) converges to 0. This means that for any small positive number ε, there exists a positive integer n0 such that for all n ≥ n0, the absolute value of a(n) is less than ε^2. We can rewrite this as: for any small positive number ε, there exists a positive integer n0 such that for all n ≥ n0, the absolute value of a(n) is less than ε^2.

Since we are trying to prove that √a(n) converges to 0, we can use the definition of convergence to rewrite this as: for any small positive number ε, there exists a positive integer n0 such that for all n ≥ n0, the absolute value of √a(n) is less than √(ε^2).

Now, we can simplify this to: for any small positive number ε, there exists a positive integer n0 such that for all n ≥ n0, the absolute value of √a(n) is less than ε.

This is exactly what we were trying to prove! We have shown that for any small positive number ε, there exists a positive integer n0 such that for all n ≥ n0, the absolute value of √a(n) is less than ε, which is the definition of convergence to 0.

Therefore, we have proven that √a(n) converges to 0. I hope this helps you with your proof.