Σ[k=1,n]1/{p+(k-1)q}(p+kq){p+(k+1)q}などの級数

\begin{alignat}{2}
&(1)  \displaystyle\sum_{k=1}^n \frac{1}{\{p+(k-1)q\}(p+kq)}=\frac{n}{p(p+nq)}\\
&(2)  \displaystyle\sum_{k=1}^n \frac{1}{\{p+(k-1)q\}(p+kq)\{p+(k+1)q\}}=\frac{n(2p+nq+q)}{2p(p+q)(p+nq)\{p+(n+1)q\}}\\
&(3)  \displaystyle\sum_{k=1}^n \frac{1}{\{p+(k-1)q\}(p+kq)\{p+(k+1)q\} \cdots \{p+(k+l)q\}}\\
&=\frac{1}{(l+1)q}\left[\frac{1}{p(p+q) \cdots (p+lq)}-\frac{1}{(p+nq)\{p+(n+1)q\} \cdots \{p+(n+l)q\}}\right]\\
\end{alignat}ただし、全て \(p,q≠0,\,l \in \mathrm{N}\)










<証明>

全て部分分数分解を行い、

シグマを効かせて始めの項と終わりの項以外を消去します。
\begin{alignat}{2}
&(1)  \displaystyle\sum_{k=1}^n \frac{1}{\{p+(k-1)q\}(p+kq)}=\frac{1}{q}\displaystyle\sum_{k=1}^n \left\{\frac{1}{p+(k-1)q}-\frac{1}{p+kq}\right\}\\
&=\frac{1}{q}\bigg[\left(\frac{1}{p}-\frac{1}{p+q}\right)+\left(\frac{1}{p+q}-\frac{1}{p+2q}\right)+\left(\frac{1}{p+2q}-\frac{1}{p+3q}\right)+ \cdots \\
&    \cdots +\left\{\frac{1}{p+(n-2)q}-\frac{1}{p+(n-1)q}\right\}+\left\{\frac{1}{p+(n-1)q}-\frac{1}{p+nq}\right\}\bigg]\\
&=\frac{1}{q}\left(\frac{1}{p}-\frac{1}{p+nq}\right)=\frac{1}{q} \cdot \frac{p+nq-p}{p(p+nq)}=\frac{n}{p(p+nq)}
\end{alignat}以上より$$\displaystyle\sum_{k=1}^n \frac{1}{\{p+(k-1)q\}(p+kq)}=\frac{n}{p(p+nq)}$$







\begin{alignat}{2}
&(2)  \displaystyle\sum_{k=1}^n \frac{1}{\{p+(k-1)q\}(p+kq)\{p+(k+1)q\}}\\
&=\displaystyle\sum_{k=1}^n \frac{1}{p+kq} \cdot \frac{1}{\{p+(k-1)q\}\{p+(k+1)q\}}\\
&=\displaystyle\sum_{k=1}^n \frac{1}{p+kq} \cdot \frac{1}{2q} \left\{\frac{1}{p+(k-1)q}-\frac{1}{p+(k+1)q}\right\}\\
&=\frac{1}{2q}\displaystyle\sum_{k=1}^n \left[ \frac{1}{\{p+(k-1)q\}(p+kq)}-\frac{1}{(p+kq)\{p+(k+1)q\}}\right]\\
&=\frac{1}{2q}\bigg[\left\{\frac{1}{p(p+q)}-\frac{1}{(p+q)(p+2q)}\right\}+\left\{\frac{1}{(p+q)(p+2q)}-\frac{1}{(p+2q)(p+3q)}\right\}+ \cdots \\
&      \cdots \left\{\frac{1}{(p+(n-2)q)(p+(n-1)q)}-\frac{1}{(p+(n-1)q)(p+nq)}\right\}+\left\{\frac{1}{(p+(n-1)q)(p+nq)}-\frac{1}{(p+nq)(p+(n+1)q)}\right\}\bigg]\\
&=\frac{1}{2q}\left[\frac{1}{p(p+q)}-\frac{1}{(p+nq)\{p+(n+1)q\}}\right]\\
&=\frac{1}{2q} \cdot \frac{(p+nq)\{p+(n+1)q\}-p(p+q)}{p(p+q)(p+nq)\{p+(n+1)q\}}\\
&=\frac{1}{2q} \cdot \frac{p^2+(n+1)pq+npq+n(n+1)q^2-p^2-pq}{p(p+q)(p+nq)\{p+(n+1)q\}}\\
&=\frac{1}{2q} \cdot \frac{2npq+n(n+1)q^2}{p(p+q)(p+nq)\{p+(n+1)q\}}\\
&=\frac{1}{2q} \cdot \frac{nq(2p+nq+q)}{p(p+q)(p+nq)\{p+(n+1)q\}}\\
&=\frac{n(2p+nq+q)}{2p(p+q)(p+nq)\{p+(n+1)q\}}
\end{alignat}以上より$$\displaystyle\sum_{k=1}^n \frac{1}{\{p+(k-1)q\}(p+kq)\{p+(k+1)q\}}=\frac{n(2p+nq+q)}{2p(p+q)(p+nq)\{p+(n+1)q\}}$$







\begin{alignat}{2}
&(3)  \displaystyle\sum_{k=1}^n \frac{1}{\{p+(k-1)q\}(p+kq)\{p+(k+1)q\} \cdots \{p+(k+l)q\}}\\
&=\displaystyle\sum_{k=1}^n \frac{1}{(p+kq)\{p+(k+1)q\} \cdots \{p+(k+l-1)q\}}\cdot \frac{1}{\{p+(k-1)q\}\{p+(k+l)q\}}\\

&=\displaystyle\sum_{k=1}^n \frac{1}{(p+kq)\{p+(k+1)q\} \cdots \{p+(k+l-1)q\}}\cdot\frac{1}{(l+1)q}\left[\frac{1}{p+(k-1)q}-\frac{1}{p+(k+l)q}\right]\\
&=\frac{1}{(l+1)q}\displaystyle\sum_{k=1}^n\left[\frac{1}{\{p+(k-1)q\}(p+kq)\cdots \{p+(k+l-1)q\}}-\frac{1}{(p+kq)\{p+(k+1)q\} \cdots \{p+(k+l)q\}}\right]\\
&=\frac{1}{(l+1)q} \bigg[\left\{\frac{1}{p(p+q) \cdots (p+lq)}-\frac{1}{(p+q)(p+2q)\cdots (p+(l+1)q)}\right\}\\
&  +\left\{\frac{1}{(p+q)(p+2q) \cdots (p+(l+1)q)}-\frac{1}{(p+2q)(p+3q)\cdots (p+(l+2)q)}\right\}+ \cdots \\
&     +\cdots \left\{\frac{1}{(p+(n-1)q)(p+nq) \cdots (p+(n+l-1)q)}-\frac{1}{(p+nq)(p+(n+1)q)\cdots (p+(n+l)q)}\right\}\bigg]\\
&=\frac{1}{(l+1)q}\left[\frac{1}{p(p+q) \cdots (p+lq)}-\frac{1}{(p+nq)\{p+(n+1)q\} \cdots \{p+(n+l)q\}}\right]
\end{alignat}以上より$$\displaystyle\sum_{k=1}^n \frac{1}{\{p+(k-1)q\}(p+kq)\{p+(k+1)q\} \cdots \{p+(k+l)q\}}=\frac{1}{(l+1)q}\left[\frac{1}{p(p+q) \cdots (p+lq)}-\frac{1}{(p+nq)\{p+(n+1)q\} \cdots \{p+(n+l)q\}}\right]$$







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