求证：$\displaystyle \sum\limits_{k=1}^n {\dfrac{1}{{{2^k}-{{\left( {-1} \right)}^k}}}}<\dfrac{{11}}{{12}}$．
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【答案】
略
【解析】
此时不能直接令 $q=\dfrac{1}{2}$ 进行等比放缩，而需要对交错项进行恰当的处理．
直接放缩掉负项 $$\sum\limits_{k=1}^n {\dfrac{1}{{{2^k}-{{\left( {-1} \right)}^k}}}}<\dfrac{1}{{{2^1}+1}}+\dfrac{1}{{{2^2}-1}}+\dfrac{1}{{{2^3}+1}}+\sum\limits_{k=4}^n {\dfrac{1}{{{2^k}-1}}}
<\dfrac{7}{9}+\dfrac{{\dfrac{1}{{{2^4}-1}}}}{{1-\dfrac{1}{2}}}
=\dfrac{{41}}{{45}}<\dfrac{{11}}{{12}}.$$分为两个子列 $$\sum\limits_{k=1}^{2n} {\dfrac{1}{{{2^k}-{{\left( {-1} \right)}^k}}}}=\sum\limits_{k=1}^n {\dfrac{1}{{{4^k}-1}}} {\rm{+}}\sum\limits_{k=1}^n {\dfrac{1}{{2 \cdot {4^{k-1}}+1}}} .$$其中$$\sum\limits_{k=1}^n {\dfrac{1}{{{4^k}-1}}}=\dfrac{1}{{{4^1}-1}}+\sum\limits_{k=2}^n {\dfrac{1}{{{4^k}-1}}}<\dfrac{1}{3}+\dfrac{{\dfrac{1}{{{4^2}-1}}}}{{1-\dfrac{1}{4}}}= \dfrac{{19}}{{45}},$$$$\sum\limits_{k=1}^n {\dfrac{1}{{2 \cdot {4^{k-1}}+1}}}<\dfrac{1}{3}+\dfrac{1}{9}+\sum\limits_{k=3}^n {\dfrac{1}{{2 \cdot {4^{k-1}}}}}<\dfrac{4}{9}+\dfrac{{\dfrac{1}{{32}}}}{{1-\dfrac{1}{4}}}=\dfrac{{35}}{{72}},$$此时$$\dfrac{{19}}{{45}}+\dfrac{{35}}{{72}}<\dfrac{{11}}{{12}},$$因此原命题得证．
将交错项分别合并 $$\sum\limits_{k=1}^{2n} {\dfrac{1}{{{2^k}-{{\left( {-1} \right)}^k}}}}=\sum\limits_{k=1}^n {\left( {\dfrac{1}{{{4^k}-1}}+\dfrac{1}{{2 \cdot {4^{k-1}}+1}}} \right)}=\sum\limits_{k=1}^n {\dfrac{{3 \cdot {4^k}}}{{\left( {{4^k}-1} \right)\left( {{4^k}+2} \right)}}},$$可以放缩为等比数列$$\sum\limits_{k=1}^n {\dfrac{{3 \cdot {4^k}}}{{\left( {{4^k}-1} \right)\left( {{4^k}+2} \right)}}}=3 \cdot \sum\limits_{k=1}^n {\dfrac{1}{{{4^k}-\dfrac{2}{{{4^k}}}+1}}}<3 \cdot \left( {\dfrac{1}{{4-\dfrac{2}{4}+1}}+\sum\limits_{k=2}^n {\dfrac{1}{{{4^k}}}} } \right)< 3\left( {\dfrac{2}{9}+\dfrac{{\dfrac{1}{{16}}}}{{1-\dfrac{1}{4}}}} \right)=\dfrac{{11}}{{12}}.$$
<\dfrac{7}{9}+\dfrac{{\dfrac{1}{{{2^4}-1}}}}{{1-\dfrac{1}{2}}}
=\dfrac{{41}}{{45}}<\dfrac{{11}}{{12}}.$$
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