Express $S =\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\ldots+\frac{1}{(n-2)\cdot(n-1)}+\frac{1}{n\cdot(n+1)}$ in terms of $n$. 
Express $S =\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\ldots+\frac{1}{(n-2)\cdot(n-1)}+\frac{1}{n\cdot(n+1)}$ in terms of $n$.

Here's what I have done so far:
$$\begin{align*}
S &=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{n-2}-\frac{1}{n-1}\right)+\left(\frac{1}{n}-\frac{1}{n+1}\right)\\[5pt]
&= \left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\ldots+\frac{1}{n}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots+\frac{1}{n+1}\right)\\[5pt]
&= \sum_{k=1}^n \frac{1}{2k-1} - \sum_{k=1}^{n+1} \frac{1}{2k} \\[5pt]
\end{align*}$$
How do I continue the problem from here? Is it even possible?
 A: It seems the expression $S$ needs to be revised somewhat. If we take a look at the first two summands
\begin{align*}
\frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\cdots
\end{align*}
we see the left-hand factor in the denominator is odd and the right-hand factor is even. Taking $n$ summands of this kind we have
\begin{align*}
S_{1}=\frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+\cdots+\frac{1}{(2n-1)2n}\tag{1}
\end{align*}
where the last summand has an odd and even factor of the given form in the denominator.

Here we calculate $S_1$. We obtain from (1)
\begin{align*}
\color{blue}{S_1=\sum_{k=1}^n\frac{1}{(2k-1)2k}}
&=\sum_{k=1}^n\left(\frac{1}{2k-1}-\frac{1}{2k}\right)\\
&=\sum_{k=1}^n\frac{1}{2k-1}-\frac{1}{2}\sum_{k=1}^n\frac{1}{k}\tag{2}\\
&=\left(\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^{n}\frac{1}{2k}\right)-\frac{1}{2}\sum_{k=1}^n\frac{1}{k}\tag{3}\\
&=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^n\frac{1}{k}\tag{4}\\
&\,\,\color{blue}{=H_{2n}-H_{n}}\tag{5}
\end{align*}

Comment:

*

*In (2) we split the sum into two sums and factor out the constant $\frac{1}{2}$ from the right-hand sum.


*In (3) we add and subtract summands with even denominator to the left-hand sum for convenient calculations in the further steps. This does not change the value of the sum since we are adding zero only.


*In (4) we simplify and collect the sums.


*In (5) we write the sums using the symbol for Harmonic numbers.
A: Let $ n\in\mathbb{N}^{*} $ be an odd integer, we have : \begin{aligned}\frac{1}{1\times 2}+\frac{1}{3\times 4}+\cdots +\frac{1}{n\left(n+1\right)}=\sum_{k=1}^{\frac{n+1}{2}}{\frac{1}{2k\left(2k-1\right)}}&=\sum_{k=1}^{\frac{n+1}{2}}{\frac{1}{2k-1}}-\sum_{k=1}^{\frac{n+1}{2}}{\frac{1}{2k}}\\ &=\sum_{k=1}^{n}{\frac{1}{k}}-\sum_{k=1}^{\frac{n-1}{2}}{\frac{1}{2k}}-\sum_{k=1}^{\frac{n+1}{2}}{\frac{1}{2k}}\\ \sum_{k=1}^{\frac{n+1}{2}}{\frac{1}{2k\left(2k-1\right)}}&=H_{n}-\frac{H_{\frac{n-1}{2}}+H_{\frac{n+1}{2}}}{2}\end{aligned}
A: So , we need to solve for : $$\sum_{k=1}^{\frac{n+1}{2}}\frac{1}{2k-1}-\sum_{k=1}^{\frac{n+1}{2}}\frac{1}{2k}=\psi$$ Which is also , $$\psi=\left(1+\frac{1}{3}+\frac{1}{5}+........+\frac{1}{n}\right)-\frac{1}{2}S_{n+1}$$ $$\psi=\left(1+\frac{1}{2}+\frac{1}{3}+........+\frac{1}{n}+\frac{1}{n+1}\right)-S_{n+1}$$ Hence , $$\psi=0$$
