Proving limits for converging sequences

[Ed Quanta]

If a general statement like an->a where (an) is a sequence of non negative real numbers, how would we prove the sqare root an->the square root of a.

When a=0, this can easily be done. But I don't see how this is possible from the given information for the case where a>0. Thanks for any help or suggestions.

 

[courtrigrad]

Hello

Use the fact that ( 1+ h ) ^ n > 1 + nh (corollary). We know that the sequence (an) is non negative, hence 1 + h is positive. Work of this, and you should be able to get the desired result. Try using this corollary for an = (nth rooth of n)

 

[M98Ranger]

To prove the limit for the case where a>0, we can use the definition of a limit for a converging sequence. According to the definition, for any positive number ε, there exists a natural number N such that for all n≥N, the absolute difference between an and a is less than ε. In other words, |an-a|<ε.

Now, let's consider the sequence (√an) where an is a sequence of non-negative real numbers. We want to show that (√an) converges to √a. So, for any positive number ε, we need to find a natural number N such that for all n≥N, |√an-√a|<ε.

First, we can rewrite |√an-√a| as |√an-√a|/|√an+√a| * |√an+√a|. Since an>0, we know that √an+√a>0. So, we can choose N such that for all n≥N, |√an-√a|/|√an+√a|<ε/√an+√a. This means that |√an-√a|<ε/√an+√a * |√an+√a| = ε.

Hence, for any positive number ε, we have found a natural number N such that for all n≥N, |√an-√a|<ε. This satisfies the definition of a limit and proves that (√an) converges to √a.

In summary, to prove the limit for a converging sequence (√an), we need to use the definition of a limit and manipulate the expression to show that it satisfies the definition. In this case, we used the fact that (√an+√a)>0 to simplify the expression and find a suitable N.