Proving convergence by definition

Question

I need to use definition of convergence to prove that this sequence converges. $$ a_n=\sqrt{(n^2+n)}-\sqrt{(n^2-n)}$$ I know it converges to 1, so I start with $$(\forall\epsilon>0)(\exists N \in \Bbb N)(\forall n >N) |a_n-1|<\epsilon$$ $$|\sqrt{n^2+n}-\sqrt{n^2-n}-1|$$ I tried multiplying with $$\frac{\sqrt{n^2+n}+\sqrt{n^2-n}}{\sqrt{n^2+n}+\sqrt{n^2-n}}$$ but it leads me nowhere. I know in the end I should compare it with some sequence that converges to zero, but I can't get to there.

Answer 1

you can write $$\sqrt{n^2+n}-(\sqrt{n^2-n}+1)$$ and multiply by $$\frac{\sqrt{n^2+n}+\sqrt{n^2-n}+1}{\sqrt{n^2+n}+\sqrt{n^2-n}+1}$$

Answer 2

Let $b_n=\sqrt{n^2+n}$. Note that $a_n=b_n-b_{n-1}$.

For $0<q\le 1$, consider the expression $$\tag1\left(n+\frac q2\right)^2=n^2+ qn+\frac{q^2}4.$$ With $q=1$, the right hand side is $>n^2+ n$, hence $$\tag2b_n<n+\frac12.$$ With $q=1-\frac1{4n}$, the right hand side in $(1)$ becomes $n^2+ n+\frac{q^2-1}4<n^2+ n$, hence $$b_n>n+\frac12 -\frac1{8n}.$$ We conclude $$ a_n=b_n-b_{n-1}\begin{cases}<\left(n+\frac12\right)-\left(n-1+\frac12-\frac1{8(n-1)}\right)&=1+\frac1{8(n-1)}\\ >\left(n+\frac12-\frac1{8n}\right)-\left(n-1+\frac12\right)&=1-\frac1{8n}\end{cases}$$ and so for $\epsilon>0$ we have $|a_n-1|<\epsilon $ as soon as $n>\frac 8\epsilon+1$.

Answer 3

What you tried is exactly what we need here. It gives $$a_n={2n\over\sqrt{n^2+n}+\sqrt{n^2-n}}={2\over\sqrt{1+{1\over n}}+\sqrt{1-{1\over n}}}\to1\qquad(n\to\infty)\ .$$