Convergence of Series, problems with intermediate steps 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
 A: 1) The author uses a trick to compute the exact value of the infinite product but one does not need it to show the convergence only. 
Namely, use that  $\frac{k^3-1}{k^3+1}=1-\frac2{k^3+1}$, that $\frac2{k^3+1}<1$ for every $k\ge2$ and that the series $\sum\limits_{k}\frac2{k^3+1}$ converges absolutely.
2) Use the trick of the conjugate quantity, that is,  $\sqrt{x}-\sqrt{y}=\frac{x-y}{\sqrt{x}+\sqrt{y}}$ for positive $x$ and $y$.
Thus, $a_n=\frac{n}{\sqrt{n^2+n}+n}$ (choose $x=n^2+n$ and $y=n^2$) hence $1-2a_n=\frac{\sqrt{n^2+n}-n}{\sqrt{n^2+n}+n}$, which is also $1-2a_n=\frac{n}{(\sqrt{n^2+n}+n)^2}$ (same $x$ and $y$).
Since $\sqrt{n^2+n}+n\ge2n$, this shows that $0\le 1-2a_n\le\frac1{4n}$, hence $2a_n\to1$ and, as the author writes, $\frac12-\frac1{8n}\le a_n\le\frac12$.
A: 1) Factoring further
$$
\begin{align}
k^3-1&=(k-1)(k^2+k+1)=(k-1)(k+\alpha)(k+\bar{\alpha})\\
k^3+1&=(k+1)(k^2-k+1)=(k+1)(k-\alpha)(k-\bar{\alpha})
\end{align}
$$
where $\alpha+\bar{\alpha}=1$. It is easy to see that $\prod\frac{k-1}{k+1}$ is telescoping. However, when we write $\prod\frac{(k+\alpha)(k+\bar{\alpha})}{(k-\alpha)(k-\bar{\alpha})}=\prod\frac{(k+1-\bar{\alpha})(k+1-\alpha)}{(k-\alpha)(k-\bar{\alpha})}$, it becomes evident that this telescopes, too.
2)
$$
\begin{align}
\left|\frac{n}{\sqrt{n^2+n}+n}-\frac{1}{2}\right|
&=\frac{1}{2}\left|\frac{n-\sqrt{n^2+n}}{n+\sqrt{n^2+n}}\right|\\
&=\frac{1}{2}\frac{n}{(n+\sqrt{n^2+n})^2}\\
&<\frac{1}{2}\frac{n}{4n^2}\\
&=\frac{1}{8n}
\end{align}
$$
