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?
 A: $$\frac1{n+1}=\int_{n+1}^{\infty}\frac{dx}{x^2}\leq A_n:=\sum_{k=n+1}^{\infty} \frac1{k^2}\leq\int_{n}^\infty \frac{dx}{x^2}=\frac1n$$ and similarly, $$\frac1{4n+2}\leq B_n:=\sum_{k=n+1}^{\infty}\frac1{(2k-1)^2}\leq \frac1{4n-2}$$
We see by the squeeze theorem $nA_n\to 1,$ and $nB_n\to \frac14$ and $nA_nB_n\to 0.$
Now $$\frac{\pi^4}{48}-a_nb_n=\frac{\pi^2}8A_n+\frac{\pi^2}6B_n-A_nB_n.$$
Multiplying both sided by $n$ we get a limit of $$\frac{\pi^2}8+\frac{\pi^2}{24}=\frac{\pi^2}6.$$

In general, if $f(n)(a-a_n)\to A$ and $f(n)(b-b_n)\to B$ where $f(n)\to\infty,$ then:
$$f(n)(ab-a_nb_n)=bf(n)(a-a_n)+a_nf(n)(b-b_n)\to bA+aB.$$

That formula, $bA+aB$ might look a little like the derivative of a product rule.
And it is related. If $f$ is one-to-one and never zero, we can define $a_f$ as a function on $X_f=\{0\}\cup\left\{\frac{1}{f(n)}\mid n\in\mathbb N\right\}$ with $ a_f(1/f(n))=a_n$ and $a_f(0)=a.$
Then $$A=\lim_{h\in X_f\to 0}\frac{a_f(0)-a_f(h)}{h}.$$ So $-A$ is a kind of derivative, $a_f'(0).$
Then we are saying $$(a_fb_f)'(0)=a_f'(0) b_f(0)+b_f'(0)a_f(0).$$
So we can think of $f$ as helping us define a notion of a derivative of a convergent infinite sequence at infinity. We get different derivatives at infinity depending on $f(n).$
The simplest case, when $a_n$ is increasing, is to define $f(n)=\frac{1}{a-a_n}.$ Then $a_f'(0)=1.$
A: 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.
