Let (an) be the sequence given byan = 1√2n + 1 − 1√2nProve from first principles that (an) is Cauchy.
Question
Let (an) be the sequence given byan = 1√2n + 1 − 1√2nProve from first principles that (an) is Cauchy.
Solution
To prove that a sequence (an) is Cauchy, we need to show that for any given positive number ε, there exists a positive integer N such that for all integers m, n greater than N, the absolute difference between am and an is less than ε.
Given the sequence an = 1/√(2n + 1) - 1/√(2n), let's simplify it first:
an = [√(2n) - √(2n + 1)] / [√(2n) * √(2n + 1)] = √(2n) / [√(2n) * √(2n + 1)] - √(2n + 1) / [√(2n) * √(2n + 1)] = 1 / √(2n + 1) - 1 / √(2n)
Now, let's take two terms am and an (m > n) from the sequence:
|am - an| = |1 / √(2m + 1) - 1 / √(2m) - [1 / √(2n + 1) - 1 / √(2n)]| = |1 / √(2m + 1) - 1 / √(2n + 1)| + |1 / √(2m) - 1 / √(2n)|
Since 1 / √x is a decreasing function for x > 0, we have:
1 / √(2m + 1) - 1 / √(2n + 1) < 1 / √(2n + 1) - 1 / √(2n + 2) 1 / √(2m) - 1 / √(2n) < 1 / √(2n) - 1 / √(2n + 1)
Adding these two inequalities, we get:
|am - an| < 2 / √(2n + 1) - 2 / √(2n + 2)
Now, for any given ε > 0, we can find a positive integer N such that 2 / √(2N + 1) - 2 / √(2N + 2) < ε. Therefore, for all m, n > N, |am - an| < ε, which proves that the sequence (an) is Cauchy.
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