已知 $n\in\mathbb N^{\ast}$，$q\geqslant 2$，求证：$\displaystyle \sum_{k=1}^n\dfrac{1}{kq^k}<\dfrac{7}{7q-4}$．
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【答案】
略
【解析】
考虑到\[\begin{split}\dfrac{1}{kq^k}&=\left[\dfrac{1}{(k+m)q^k}-\dfrac{1}{(k+m+1)q^{k+1}}\right]\cdot \dfrac{(k+m)(k+m+1)}{k\left(k+m+1-\dfrac{k+m}q\right)}\\
&\leqslant \left[\dfrac{1}{(k+m)q^k}-\dfrac{1}{(k+m+1)q^{k+1}}\right]\cdot \dfrac{q}{q-1}\cdot \dfrac{k^2+(2m+1)k+m^2+m}{k^2+\left(m+\dfrac{q}{q-1}\right)k},\end{split}\]令\[2m+1=m+\dfrac{q}{q-1}\]即\[m=\dfrac{1}{q-1},\]于是有\[\begin{split}\sum_{k=p}^{n}\dfrac{1}{kq^k}&<\dfrac{1}{\left(p+\dfrac{1}{q-1}\right)q^p}\cdot \dfrac{q}{q-1}\cdot \left[1+\dfrac{q}{(q-1)^2p^2+(q^2-1)p}\right]\\
&=\dfrac{(p+1)q-p}{p(q-1)[(p+1)q-p+1]\cdot q^{p-1}},\end{split}\]令 $p=1$，可得\[\sum_{k=1}^n\dfrac{1}{kq^k}<\dfrac{2q-1}{2q(q-1)},\]而\[\dfrac{7}{7q-4}-\dfrac{2q-1}{2q(q-1)}=\dfrac{q-4}{2q(q-1)(7q-4)},\]不符合题意．
令 $p=2$，可得\[\sum_{k=1}^n\dfrac{1}{kq^k}<\dfrac 1q+\dfrac{3q-2}{2q(3q^2-4q+1)}=\dfrac{6q-5}{2(3q^2-4q+1)},\]而\[\dfrac{7}{7q-4}-\dfrac{6q-5}{2(3q^2-4q+1)}=\dfrac{3(q-2)}{2(7q-4)(3q^2-4q+1)}\geqslant 0,\]于是原命题得证．
&\leqslant \left[\dfrac{1}{(k+m)q^k}-\dfrac{1}{(k+m+1)q^{k+1}}\right]\cdot \dfrac{q}{q-1}\cdot \dfrac{k^2+(2m+1)k+m^2+m}{k^2+\left(m+\dfrac{q}{q-1}\right)k},\end{split}\]令\[2m+1=m+\dfrac{q}{q-1}\]即\[m=\dfrac{1}{q-1},\]于是有\[\begin{split}\sum_{k=p}^{n}\dfrac{1}{kq^k}&<\dfrac{1}{\left(p+\dfrac{1}{q-1}\right)q^p}\cdot \dfrac{q}{q-1}\cdot \left[1+\dfrac{q}{(q-1)^2p^2+(q^2-1)p}\right]\\
&=\dfrac{(p+1)q-p}{p(q-1)[(p+1)q-p+1]\cdot q^{p-1}},\end{split}\]令 $p=1$，可得\[\sum_{k=1}^n\dfrac{1}{kq^k}<\dfrac{2q-1}{2q(q-1)},\]而\[\dfrac{7}{7q-4}-\dfrac{2q-1}{2q(q-1)}=\dfrac{q-4}{2q(q-1)(7q-4)},\]不符合题意．
令 $p=2$，可得\[\sum_{k=1}^n\dfrac{1}{kq^k}<\dfrac 1q+\dfrac{3q-2}{2q(3q^2-4q+1)}=\dfrac{6q-5}{2(3q^2-4q+1)},\]而\[\dfrac{7}{7q-4}-\dfrac{6q-5}{2(3q^2-4q+1)}=\dfrac{3(q-2)}{2(7q-4)(3q^2-4q+1)}\geqslant 0,\]于是原命题得证．
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