<s>[INST] 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.
 [/INST] 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
$$\sum_{k=1}^n\frac{1}{2k(2k-1)}=\sum_{k=1}^n\left(\frac{1}{2k-1}+\frac{1}{2k}\right)-\sum_{k=1}^n\frac{1}{k}=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^n\frac{1}{k}=\sum_{k=n+1}^{2n}\frac{1}{k}$$
</s>