リーマン和
[ 1 , 2 ] {\displaystyle [1,2]} で f ( x ) = 1 x {\displaystyle f(x)={\frac {1}{x}}} のとき
等比数列 x k = 2 k n ( k = 0 , 1 , 2 , ... , n ) {\displaystyle x_{k}=2^{\frac {k}{n}}\;(k=0,1,2,\dots ,n)} をとると、 左リーマン和と右リーマン和は、それぞれ、
∑ k = 1 n 1 x k − 1 Δ x k = ∑ k = 1 n 1 2 k − 1 n ( 2 k n − 2 k − 1 n ) = n ( 2 1 n − 1 ) {\displaystyle \sum _{k=1}^{n}{\frac {1}{x_{k-1}}}\Delta x_{k}=\sum _{k=1}^{n}{\frac {1}{2^{\frac {k-1}{n}}}}\,(2^{\frac {k}{n}}-2^{\frac {k-1}{n}})=n\left(2^{\frac {1}{n}}-1\right)}
∑ k = 1 n 1 x k Δ x k = ∑ k = 1 n 1 2 k n ( 2 k n − 2 k − 1 n ) = n ( 1 − 2 − 1 n ) {\displaystyle \sum _{k=1}^{n}{\frac {1}{x_{k}}}\Delta x_{k}=\sum _{k=1}^{n}{\frac {1}{2^{\frac {k}{n}}}}\,(2^{\frac {k}{n}}-2^{\frac {k-1}{n}})=n\left(1-2^{-{\frac {1}{n}}}\right)}