I have problems with two exercises. I know the answer of those two but both of them have a step which I don't understand.
1) I have to prove that $\prod_{k=2}^n \frac{k^3-1}{k^3+1}$ is convergent.
My first steps were:
$\prod_{k=2}^n \frac{k^3-1}{k^3+1}= \prod_{k=2}^n \frac{k-1}{k+1}\cdot \prod_{k=2}^n \frac{k^2+k+1}{k^2-k+1}=\frac{2(n-1)!}{(n+1)!}\prod_{k=2}^n \frac{k^2+k+1}{k^2-k+1}$
And then I got stuck. The author of the exercise does that:
$\frac{2(n-1)!}{(n+1)!}\cdot\prod_{k=2}^n \frac{k^2+k+1}{k^2-k+1} = \frac{2}{n(n+1)}\cdot\prod_{k=2}^n((k+1)^2-(k+1)+1)\cdot\prod_{k=2}^n\frac{1}{k^2-k+1}=$ $\frac{2}{n(n+1)}\cdot\prod_{k=3}^{n+1}(k^2-k+1)\cdot\prod_{k=2}^n\frac{1}{k^2-k+1}=\frac{2}{n(n+1)}\cdot\frac{(n+1)^2-(n+1)+1}{3}=\frac{2}{3}\cdot\frac{n^2+n+1}{n^2+n}\rightarrow\frac{2}{3}$
I understand what he does, but I don't know how he knew it. How can I use it for other exercises, so does somebody know how to remember this trick and for which kind of exercises this works?
2) I have to prove that $a_n:=\sqrt{n^2+n}-n$ is convergent and that a constant $A$ exist with $|a_n-a|<\frac{A}{n}.$
I proved that $a_n$ is convergent to $\frac{1}{2}$. For finding the constant $A$ the author says
It's easy to see, that: $|a_n -\frac{1}{2}|=|\frac{n}{\sqrt{n^2+n}+n}-\frac{1}{2}|<\frac{1}{8n}$, therefore $A:=\frac{1}{8n}$.
Unfortunately I don't see it. Can somebody please give a hint how he gets $\frac{1}{8}$?
thanks