Prove that sequence converges to zero?
Consider the sequence $$\sqrt{n+1}-\sqrt{n}$$
Let $\epsilon\gt0$. The Archimedian Property of Reals gives $n_0\in\Bbb N$ with $n_0\gt1/(4\epsilon^2)$. (Where does this come from? I know the Archimedian Property as "for each $a,b$ of $\Bbb R$ there is an $n$ such that $an>b$". Where is this in the last expression?) and for each n element of N with $n\geq n_0$ $n\geq n_0\gt1/(4\epsilon^2)$ holds. Then $$0\lt\sqrt{n+1}-\sqrt n=\dfrac1{\sqrt{n+1}+\sqrt n}\lt\dfrac{1}{\sqrt{\frac1{4\epsilon^2}}+\sqrt{\frac1{4\epsilon^2}}}=\epsilon$$
should prove that the sequence converges to 0. What is the stuff under in the last fraction? Ans how does this whole thing prove that the sequence converges to 0? I only have shown that epsilon is larger than zero.
Sometimes these proofs seem so arbitrary.
$\endgroup$1 Answer
$\begingroup$There is a concrete idea behind the formal argument. A little experimentation with a calculator will show that $\sqrt{n+1}-\sqrt{n}$ is small positive when $n$ is large. We want to show that this is indeed true, and replace the somewhat vague "small" and "large" by explicit expressions.
It is useful to multiply top and missing bottom by $\sqrt{n+1}+\sqrt{n}$. We obtain $$0\lt \sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\lt \frac{1}{2\sqrt{n}}.\tag{1}$$ (There are other ways to get an explicit inequality of the above kind. Direct inspection of $\sqrt{n+1}-\sqrt{n}$ is not enough, since both parts may be large, so the size of their difference is not immediately obvious.)
It is clear that $\frac{1}{2\sqrt{n}}$ approaches $0$ as $n\to\infty$.
But if we want to be very formal, we have to show that for any positive $\epsilon$, there is an $N$ such that if $n\gt N$, then $|(\sqrt{n+1}-\sqrt{n})=0|\lt \epsilon$.
By (1), it is enough to show that there is an $N$ such that if $n\gt N$ then $\frac{1}{2\sqrt{n}}\lt \epsilon$.
Equivalently, we want to show that there is an $N$ such that if $n\gt N$, then $2\sqrt{n}\gt \frac{1}{\epsilon}$. Equivalently, we want to show that there is an $N$ such that if $n\gt N$ then $n\gt \frac{1}{4\epsilon^2}$.
Now we use the Archimedean property to assert that there really is an $N$ which is greater than $\frac{1}{4\epsilon^2}$. If we are operating at a more informal level, we take that fact for granted, and just say let $N$ be the greatest integer which is $\le 1/(4\epsilon^2)$.
$\endgroup$
