调和级数近似求和公式推导

调和级数是一个非常著名的级数,对于调和级数我们有一个近似的求和公式:
∑ i = 1 n 1 i = ln ⁡ ( n + 1 ) + γ ( γ 为 欧 拉 常 数 , γ = l i m n → ∞ ∫ 1 n 1 ⌊ x ⌋ − 1 x d x 约 等 于 0.57721566490153286060651209 , ) \sum_{i=1}^n\frac{1}{i}=\ln(n+1)+\gamma(\gamma为欧拉常数,\gamma=lim_{n\rightarrow\infty}\int_{1}^{n}\frac{1}{\lfloor x\rfloor}-\frac{1}{x}dx约等于0.57721566490153286060651209,) i=1ni1=ln(n+1)+γ(γ,γ=limn1nx1x1dx0.57721566490153286060651209,)
推导过程:
∑ i = 1 n 1 i = ∑ i = 1 n ∫ i i + 1 1 ⌊ x ⌋ d x \sum_{i=1}^n\frac{1}{i}=\sum_{i=1}^n\int_i^{i+1}\frac{1}{\lfloor x\rfloor}dx i=1ni1=i=1nii+1x1dx
= ∫ 1 n + 1 1 x + 1 ⌊ x ⌋ − 1 x d x =\int_1^{n+1}\frac{1}{x}+\frac{1}{\lfloor x\rfloor}-\frac{1}{x}dx =1n+1x1+x1x1dx
= ∫ 1 n + 1 1 x d x + ∫ 1 n + 1 1 ⌊ x ⌋ − 1 x d x =\int_1^{n+1}\frac{1}{x}dx+\int_1^{n+1}\frac{1}{\lfloor x\rfloor}-\frac{1}{x}dx =1n+1x1dx+1n+1x1x1dx
≈ ln ⁡ ( n + 1 ) + γ \approx\ln (n+1)+\gamma ln(n+1)+γ

你可能感兴趣的:(数学)