\begin{equation}\sum\limits_{k=1}^{n}{\frac{1}{k\left( {k+1} \right)}} = \frac{n}{n+1}\end{equation}
\begin{equation} \sum\limits_{k=1}^{n}\frac{\left \lfloor {k^\frac{1}{p}} \right\rfloor}{k(k+1)}= \sum\limits_{k=1}^{\left\lfloor {n^\frac{1}{p}} \right\rfloor}\frac{1}{{k}^{p}} \,- \frac{\left\lfloor {n^\frac{1}{p}} \right\rfloor}{n+1}, \quad p\in\mathbb{R},\,p\ge{1}\end{equation}
\begin{equation} \sum\limits_{k=1}^{n}\frac{\left \lfloor {\log_p(k)} \right\rfloor}{k(k+1)}= \frac{1 – p^{\left \lfloor {\log_p(n)} \right\rfloor}}{p-1} -\frac{\left \lfloor {\log_p(n)} \right\rfloor}{n+1}, \quad p\in\mathbb{R},\,p\gt{1}\end{equation}
\begin{equation} \sum\limits_{k=1}^{n}\frac{\left \lfloor {\log_p(k)} \right\rfloor}{4k^2-1}= \frac{1 – p^{\left \lfloor {\log_p(n)} \right\rfloor}}{4(p-1)} -\frac{\left \lfloor {\log_p(n)} \right\rfloor}{2(2n+1)}, \quad p\in\mathbb{R},\,p\gt{1}\end{equation}
\begin{equation}\sum\limits_{{k=1}}^{n}{{\frac{1}{{k\left( {k+1} \right)\left( {k+2} \right)}}}}=\frac{{n\left( {n+3} \right)}}{{4\left( {n+1} \right)\left( {n+2} \right)}}\end{equation}
\begin{align} \frac{1}{p+1}\sum\limits_{k=1}^{n}{\frac{1}{\binom{p+k}{p+1}}} &=\sum\limits_{k=1}^{n}{\frac{p\,!}{k(k+1)(k+2)\cdots (k+p)}} \\ &= \sum\limits_{l=1}^{p}{\left({\frac{1}{l}\sum\limits_{j=0}^{l-1}{\binom{p}{j}(-1)^j}} + {\frac{1}{n+l}\sum\limits_{j=l}^{p}{\binom{p}{j}(-1)^j}}\right)} , \quad p\in \mathbb{N}, \, n \ge p \\ &= \sum\limits_{l=1}^{n}{\frac{1}{l}\sum\limits_{j=0}^{l-1}{\binom{p}{j}(-1)^j}} + \sum\limits_{l=1}^{p}{\frac{1}{n+l}\sum\limits_{j=l}^{n+l-1}{\binom{p}{j}(-1)^j}} , \quad p\in \mathbb{N}, \, n \lt p \end{align}
\begin{equation}\sum\limits_{{k=1}}^{n}{{\frac{1}{{\left( {2k-1} \right)\left( {2k+1} \right)}}}}=\frac{n}{{2n+1}}\end{equation}
\begin{equation}\sum\limits_{{k=1}}^{n}{{\frac{1}{{\left( {\alpha \left( {k-1} \right)+\beta } \right)\left( {\alpha k+\beta } \right)}}}}=\frac{n}{{\beta \left( {\alpha n+\beta } \right)}}\end{equation}
\begin{align}\sum\limits_{{1\le {{i}_{1}}<{{i}_{2}}<{{i}_{3}}<\cdots <{{i}_{{n-2}}}<{{i}_{{n-1}}}\le n}} {\frac{1}{{{i}_{1}}\cdot {{i}_{2}}\cdot {{i}_{3}} \cdots {{i}_{{n-2}}}\cdot {{i}_{n-1}}}}&=\frac{1}{n\,!}\sum\limits_{1\le {i}\le n}{i}\\&=\frac{n+1}{2\left( {n-1} \right)\,!}\end{align}