User:Saiyr

$$\Pi_{i=2}^{n}(1-\frac{1}{n^{2}})=\frac{n+1}{2n}$$

$$\frac{n+1}{2n}(1-\frac{1}{(n+1)^{2}})$$

$$\frac{n+1}{2n}-\frac{n+1}{2n(n+1)^{2}}=\frac{(n+1)^{3}-(n+1)}{2n(n+1)^{2}}=\frac{(n+1)^{2}-1)}{2n(n+1)}=\frac{n^{2}+2n}{2n(n+1)}=\frac{(n+1)+1}{2(n+1}$$

$$\sum_{i=1}^{k}\sum_{j=1}^{2^{k-i+1}}1=2^{k}\sum_{i=1}^{k}\frac{1}{2^{i-1}}=2^{k}\sum_{i=0}^{k-1}\frac{1}{2^{i}}$$

$$\frac{1}{x-5} + \frac{-x-5}{x^2} + 64$$

$$\sigma_{ch-out \ne \omega}(Guest)$$