Let (an) be the sequence given byan = 1√2n + 1 − 1√2nProve from first principles that (an) is Cauchy.

To prove that the sequence $(a_n)$ is Cauchy, we need to demonstrate that for every $\epsilon > 0$, there exists an integer $N$ such that for all $m, n \geq N$, the following holds:

$|a_n - a_m| < \epsilon$

Given that:

$a_n = \frac{1}{\sqrt{2n + 1}} - \frac{1}{\sqrt{2n}}$

Step 1: Break Down the Problem

1. We will start by expressing $|a_n - a_m|$.
2. We will manipulate the expression to find an upper bound for $|a_n - a_m|$.

Step 2: Relevant Concepts

We will use the properties of absolute values and the definitions of limits and convergence.

Step 3: Analysis and Detail

1. Expressing $|a_n - a_m|$:

   We have:

   $|a_n - a_m| = \left| \left( \frac{1}{\sqrt{2n + 1}} - \frac{1}{\sqrt{2n}} \right) - \left( \frac{1}{\sqrt{2m + 1}} - \frac{1}{\sqrt{2m}} \right) \right|$

2. Combining the terms:

   Let's simplify:

   $|a_n - a_m| = \left| \left( \frac{1}{\sqrt{2n + 1}} - \frac{1}{\sqrt{2m + 1}} \right) + \left( \frac{1}{\sqrt{2m}} - \frac{1}{\sqrt{2n}} \right) \right|$

   We can use the triangle inequality:

   $|a_n - a_m| \leq \left| \frac{1}{\sqrt{2n + 1}} - \frac{1}{\sqrt{2m + 1}} \right| + \left| \frac{1}{\sqrt{2m}} - \frac{1}{\sqrt{2n}} \right|$

3. Using the difference of squares:

   For $\frac{1}{\sqrt{2n + 1}} - \frac{1}{\sqrt{2m + 1}}$:

   $\left| \frac{1}{\sqrt{2n + 1}} - \frac{1}{\sqrt{2m + 1}} \right| = \frac{|(\sqrt{2m + 1} - \sqrt{2n + 1})|}{\sqrt{(2n + 1)(2m + 1)}}$

   Similarly for $\frac{1}{\sqrt{2m}} - \frac{1}{\sqrt{2n}}$.

4. Simplifying further:

   As $n, m$ grow large, we find that:

   $\sqrt{2n + 1} \sim \sqrt{2n}, \text{ and so } \frac{1}{\sqrt{2n + 1}} \to \frac{1}{\sqrt{2n}}$

Step 4: Verify and Summarize

As $n$ and $m$ approach infinity, both differences effectively shrink due to the $\frac{1}{\sqrt{n}}$ term diminishing:

Thus:

$|a_n - a_m| \to 0$

for sufficiently large $n$ and $m$. Hence, for a given $\epsilon > 0$, we can find $N$ such that when $n, m \geq N$:

$|a_n - a_m| < \epsilon$

Final Answer

The sequence $(a_n)$ is Cauchy.