利用积分放缩证明的一个不等式

陈洪葛 posted @ Feb 17, 2014 03:26:07 PM in 数学分析 , 958 阅读

Let $k,n\in\mathbb{N},n\geq k$, prove that
\[ \frac{(k+1)^{k+1}}{k^{k}}\sum_{t=k+1}^{n}\frac{1}{t^{2}}<e \]

证明:
\[\sum_{r=m+1}^{\infty} \frac{1}{r^2}=\int_0^1 \frac{x^m \ln x}{x-1}\,dx\]
\[\begin{aligned}\left(1+\frac{1}{k}\right)^{k+1} \sum_{t=k+1}^n \frac{k}{t^2}&<\left(1+\frac{1}{k}\right)^{k+1} \sum_{t=k+1}^{\infty} \frac{k}{t^2}\\&=\left(1+\frac{1}{k}\right)^{k+1}\int_0^1 \frac{kx^k \ln x}{x-1}\,dx\\&<\left(1+\frac{1}{k}\right)^{k+1}\int_0^1 kx^{k-\frac{1}{2}} \,dx\\&=\left(1+\frac{1}{k}\right)^{k+1}\left(\frac{2k}{2k+1}\right)\\&=\left(1+\frac{1}{k}\right)^k\frac{2k+2}{2k+1}\\&<e\end{aligned}\]
 


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