I've setup a router in my basement, and an access point in my attic floor. They are connected with a patch cable. Both have the same SSID and WPA2 security. Roaming does work. But, what I want to know is, if I need to set them on different channels, or do they need the exact same channel to share. I've read many community forums and manuals, but yet it is not clear what to do because they have different opinions.


Router: Fritz!Box 7270
Accesspoint: Netgear WN604



Question: Complete induction
Answer: For a full solution, proceed like this:  

$n=1$: $$\sum_{i=1}^1 \frac{1}{(2i-1)(2i+1)} = \frac{1}{(2-1)(2+1)} = \frac{1}{3} = \frac{1}{2 \cdot 1 +1},$$
so it holds for $n=1$.  

Assume next that it holds for some generic $n$. You need to show that then it also holds for $n+1$. As it holds for $n$, you can assume that
$$\sum_{i=1}^n \frac{1}{(2i-1)(2i+1)} = \frac{n}{2n+1}. \quad (1),$$
and want to show that
$$\sum_{i=1}^{n+1} \frac{1}{(2i-1)(2i+1)} = \frac{n+1}{2(n+1)+1}. \quad (2)$$
Then:
$$\begin{align}
\sum_{i=1}^{n+1} \frac{1}{(2i-1)(2i+1} &amp;= \sum_{i=1}^n \frac{1}{(2i-1)(2i+1)} + \frac{1}{(2(n+1)-1)(2(n+1)+1)} \\
&amp; = \frac{n}{2n+1} + \frac{1}{(2n+1)(2n+3)} \quad \text{using (1)} \\
&amp; = \frac{n(2n+3)}{(2n+1)(2n+3)} + \frac{1}{(2n+1)(2n+3)} \\
&amp; = \frac{2n^2 +3n +1}{(2n+1)(2n+3)} \\ 
&amp; = \frac{(n+1)(2n+1)}{(2n+1)(2n+3)} = \frac{n+1}{2(n+1)+1},\\
\end{align}$$
which is (2), and
 was to be shown.