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