12220. Let $a_n=\sum_{k=1}^n 1/k^2$ and $b_n=\sum_{k=1}^n 1/(2k-1)^2$. Prove
$$
\lim_{n\to\infty} n\left(\frac{b_n}{a_n}-\frac34\right)=\frac3{\pi^2}.
$$
证 由
$$
\frac14a_n+b_n=\sum_{k=1}^n\left(\frac1{(2k)^2}+\frac1{(2k+1)^2}\right)=a_{2n}
$$
得
$$
\begin{align*}
n\left(\frac{b_n}{a_n}-\frac34\right)&=n\cdot\frac{a_{2n}-a_n}{a_n}=\frac{n\sum_{k=n+1}^{2n}1/k^2}{\sum_{k=1}^n1/k^2}\\
&=\frac{n\sum_{k=1}^{n}1/(k+n)^2}{\sum_{k=1}^n1/k^2}=\frac{\sum_{k=1}^{n}1/\left(\frac kn+1\right)^2\cdot1/n}{\sum_{k=1}^n1/k^2}\\
&\to\frac{\int_0^11/(1+x)^2\mathrm dx}{\sum_{k=1}^\infty1/k^2}=\frac{1/2}{\pi^2/6}=\frac3{\pi^2}.\square
\end{align*}
$$
另见
裴礼文例5.3.28。
DOI: 10.1080/00029890.2020.1822718