A question regarding Cauchy's convergence test

Solution 1:

There's only one question in the Question, so I address it...

$ \underline{\forall \epsilon>0,\forall p\in \mathbb N}, \exists N>0:n>N \implies |a_{n+p}-a_n|<\epsilon\iff \{a_n\} \text{ has a finite limit}$

Let $\varepsilon$ and $p$ be given. Example: $\varepsilon = 1/10$ and $p = 23$. We now ask, does there exist a positive $N$ such that $|\sqrt{n+p} - \sqrt{n}| < \varepsilon$ is true for all $n > N$?

Example: Since $n+23 > n > 0$, we start with the simplification $|\sqrt{n+23} - \sqrt{n}| = \sqrt{n+23} - \sqrt{n}$ to see if we obtain an inequality of the form "$n > {}$ some constant" (or a solution that is a union of intervals, at least one of which is of that form)... \begin{align*} \sqrt{n+23} - \sqrt{n} &< \frac{1}{10} \\ |n+23| - 2 \sqrt{n+23}\sqrt{n} + |n| &< \frac{1}{100} \\ n + \frac{23}{2} - \sqrt{n^2+23n} &< \frac{1}{200} \\ - \sqrt{n^2+23n} &< \frac{1}{200} - n - \frac{23}{2} \\ \sqrt{n^2+23n} &> n + \frac{2299}{200} \\ n^2+23n &> n^2 + \frac{2299n}{100} + \frac{5\,285\,401}{40\,000} \\ 23n - \frac{2299n}{100} &> \frac{5\,285\,401}{40\,000} \\ n &> \frac{5\,285\,401}{400} \\ n &> 13\,213.5025 \text{.} \end{align*} So, if we set $N = 13\,213$ (or any larger number), then $n > N$ implies $|\sqrt{n+23} - \sqrt{n}| < 1/10$.

So let's repeat those manipulations for the general case. Since $n+p \geq n > 0$, we start with the simplification $|\sqrt{n+p} - \sqrt{n}| = \sqrt{n+p} - \sqrt{n}$... \begin{align*} \sqrt{n+p} - \sqrt{n} &< \varepsilon \\ |n+p| - 2 \sqrt{n+p}\sqrt{n} + |n| &< \varepsilon^2 \\ n + \frac{p}{2} - \sqrt{n^2+pn} &< \frac{1}{2}\varepsilon^2 \\ - \sqrt{n^2+pn} &< \frac{1}{2}\varepsilon^2 - n - \frac{p}{2} \\ \sqrt{n^2+pn} &> n + \frac{p-\varepsilon^2}{2} \\ n^2+pn &> n^2 + (p-\varepsilon^2)n + \frac{(p-\varepsilon^2)^2}{4} \\ pn - (p-\varepsilon^2)n &> \frac{(p-\varepsilon^2)^2}{4} \\ \varepsilon^2n &> \frac{(p-\varepsilon^2)^2}{4} \\ n &> \frac{(p-\varepsilon^2)^2}{4\varepsilon^2} \text{.} \end{align*} So, if we set $N = \max\left\{1,\frac{(p-\varepsilon^2)^2}{4\varepsilon^2}\right\}$ (or any larger number), then $n > N$ implies $|\sqrt{n+p} - \sqrt{n}| < \varepsilon$.

Since a suitable choice of $N$ exists, the left-hand side of the biimplication in the wrong statement is true for $a_n = \sqrt{n}$. That is, with that choice of $a_n$, $$ \forall \epsilon>0,\forall p\in \mathbb N, \exists N>0:n>N \implies |a_{n+p}-a_n|<\epsilon $$ is true; and if we accept the biimplication, we conclude (wrongly) $\{\sqrt{n}\}_{n \in \Bbb{N}}$ has a finite limit.