<s>[INST] Finding $\lim_{n \to \infty}\left ( \frac{\pi^4}{48}-a_nb_n \right )n$ Consider the sequences $a_n$ and $b_n$ such that $a_n=\sum_{k=1}^{n}\frac{1}{k^2}$ and $b_n=\sum_{k=1}^{n}\frac{1}{(2k-1)^2}$. Compute $$\lim_{n \to \infty}\left ( \frac{\pi^4}{48}-a_nb_n \right )n$$
When I saw this question, my first thought is that to use $\sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}$ and $\sum_{k=1}^{\infty}\frac{1}{(2k-1)^2}=\frac{\pi^2}{8}$
So $a_n=\frac{\pi^2}{6}-\sum_{k=n+1}^{\infty}\frac{1}{k^2}$ and $b_n=\frac{\pi^2}{8}-\sum_{k=n+1}^{\infty}\frac{1}{(2k-1)^2}$
But after substituting $a_n$ and $b_n$ into the $\lim_{n \to \infty}\left ( \frac{\pi^4}{48}-a_nb_n \right )n$, I could not proceed.
Is my approach a correct way to start or any other method to solve?
 [/INST] Hint: Notice that
\begin{align}
 (ab-a_nb_n)n = n(a-a_n)b+a_n(b-b_n)n.
\end{align}
Next, observe that
\begin{align}
n(a-a_n) = n\left(\frac{\pi^2}{6}-a_n\right) = n\sum^\infty_{k=n+1}\frac{1}{k^2}
\end{align}
and that
\begin{align}
\frac{n}{n+1}=n\int^\infty_{n+1}\frac{1}{x^2} dx\le n\sum^\infty_{k=n+1}\frac{1}{k^2}\le n\int^\infty_{n}\frac{1}{x^2} dx =1.
\end{align}
You can do the rest.
</s>