还是写一写利用那个级数结论来证明此题的过程吧:由
\[\sum_{k=1}^\infty\frac1{k^2}=\frac{\pi^2}6,\]得
\[\frac6{\pi^2}\sum_{k=1}^n{\frac1{k^2}}=\sum_{k=1}^n\frac1{k^2}\biggm/\left( \sum_{k=1}^n\frac1{k^2}+\sum_{k=n+1}^\infty\frac1{k^2} \right)=\frac1{1+p(n)},\]其中
\[p(n)=\sum_{k=n+1}^\infty\frac1{k^2}\biggm/\sum_{k=1}^n\frac1{k^2},\]则
\begin{align*}
p(n)&<\sum_{k=n+1}^\infty\frac1{k^2-1/4}\biggm/\sum_{k=1}^n\frac1{k(k+1)}=\frac{2(n+1)}{n(2n+1)},\\
p(n)&>\sum_{k=n+1}^\infty\frac1{k(k+1)}\biggm/\sum_{k=1}^n\frac1{k^2-1/4}=\frac{2n+1}{4n(n+1)},
\end{align*}代回去化简即得
\[1-\frac{2(n+1)}{2n^2+3n+2}<\frac6{\pi^2}\sum_{k=1}^n{\frac1{k^2}}<1-\frac{2n+1}{4n^2+6n+1},\]乘以 `(2n+1)^2/6` 再化简可得
\[\frac{2n^2}3-\frac16+\frac{2n+1}{3(2n^2+3n+2)}<\frac{(2n+1)^2}{\pi^2}\sum_{k=1}^n{\frac1{k^2}}<\frac{2n^2}3+\frac n3+\frac16-\frac{4n+1}{6(4n^2+6n+1)},\]显然此结果强于原题。 |