proving $\frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+\cdots+$ Without Induction i proved that:
$$
\begin{align}
& {} \quad \frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+\cdots+\frac{1}{(2n-1)\cdot 2n} \\[10pt]
& =\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+1}+\cdots+\frac{1}{2n}\text{ for }n\in \mathbb{N}
\end{align}
$$
by induction.
i wonder if it can be done without using induction. if so, i'll appreciate if someone could show how.
thanks.
 A: We have
$$\frac{1}{2k(2k-1)}=\frac{1}{2k-1}-\frac{1}{2k}=\frac{1}{2k-1}+\frac{1}{2k}-\frac{1}{k}$$
so
$$\sum_{k=1}^n\frac{1}{2k(2k-1)}=\sum_{k=1}^n\left(\frac{1}{2k-1}+\frac{1}{2k}\right)-\sum_{k=1}^n\frac{1}{k}=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^n\frac{1}{k}=\sum_{k=n+1}^{2n}\frac{1}{k}$$
A: $$\text{As }\frac1{(2r-1)2r}=\frac{2r-(2r-1)}{(2r-1)2r}=\frac1{2r-1}-\frac1{2r},$$
$$\text{we can write }\frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+...+\frac{1}{(2n-1)\cdot 2n}$$
$$=\left(\frac11-\frac12\right)+\left(\frac13-\frac14\right)+\cdots+\left(\frac1{2n-1}-\frac1{2n}\right)$$
$$=\left(\frac11+\frac12-2\cdot\frac12\right)+\left(\frac13+\frac14-2\cdot\frac14\right)+\cdots+\left(\frac1{2n-1}+\frac1{2n}-2\cdot\frac1{2n}\right)$$
$$=\frac11+\frac12+\frac13+\frac14+\cdots+\frac1{n-1}+\frac1n+\frac1{n+1}+\cdots+\frac1{2n-1}+\frac1{2n}-2\left(\frac12+\frac14+\frac16+\cdots+\frac1{2n}\right)$$
$$=\frac11+\frac12+\frac13+\frac14+\cdots+\frac1{n-1}+\frac1n+\frac1{n+1}+\cdots+\frac1{2n-1}+\frac1{2n}-\left(1+\frac12+\frac13+\cdots+\frac1n\right)$$
$$=\frac1{n+1}+\frac1{n+2}+\cdots+\frac1{2n-1}+\frac1{2n}$$
A: Let it be true for n
Then for $n+1$ we have,
$\displaystyle \frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+...+\frac{1}{(2n+1)\cdot 2(n+1)}$
$\displaystyle =(\frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+...+\frac{1}{(2n-1)\cdot 2n})+\frac{1}{(2n+1)\cdot 2(n+1)}$(Using induction hypothesis for $n$)
$\displaystyle =\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}+\frac{1}{(2n+1)\cdot 2(n+1)}$
$\displaystyle =\frac{1}{n+1}+\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2(n+1)}-\frac{1}{2n+1}-\frac{1}{2(n+1)}+\frac{1}{(2n+1)\cdot 2(n+1)}$
$\displaystyle =\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2(n+1)}-\frac{4n+3}{(2n+1)2(n+1)}+\frac{1}{(2n+1)2(n+1)}$
$\displaystyle =\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2(n+1)}-\frac{4n+2}{(2n+1)2(n+1)}$
$\displaystyle =\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2(n+1)}-\frac{1}{(n+1)}$
$\displaystyle =\frac{1}{n+2}+...+\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2(n+1)}$
