A problem in limits to show that $\lim_{n\to\infty}$ $p_n$/$q_n$ = $π/4$ Define two sequences of real numbers $p_n$ and $q_n$ with $n = −1, 0, 1, 2, \ldots$ as follows:
$$p_n = 2p_{n−1} + (2n−1)^2p_{n−2}$$ and
$$q_n = 2q_{n−1} + (2n−1)^2q_{n−2}$$
for every $n \geq 1$, and starting with $p_{−1}= 0, q_{−1}= 1, p_0 = q_0 = 1$.
Show that
$$\lim_{n \to \infty}\frac{p_n}{q_n} = \frac{\pi}{4}.$$
I did:
Suppose $a_n$ is a sequence that satisfies $a_n = 2a_{n−1} + (2n − 1)^2a_{n−2}$ for all
$n\geqslant1$. Subtracting $(2n + 1)a_{n−1}$ from both sides we derive a simple recursion for a new
sequence $b_n = a_n − (2n + 1)a_{n−1}$, namely $b_n = −(2n − 1)b_{n−1}$.
What to do after this... Is it correct?
If not kindly provide a detailed answer in the answer section.. As it helps a new beginner like me to understand who has just started to solve advanced problems.
 A: Let $r_n := p_n/q_n$. We can compute the first few times to observe that: $r_0 = 1$, $r_1 = \frac{2}{3} = 1-\frac{1}{3}$, $r_2 = \frac{13}{15} =1-\frac{1}{3} + \frac{1}{5}$, $r_3 = \frac{76}{105} = 1-\frac{1}{3} + \frac{1}{5} -\frac{1}{7}$. So this gives us enough motivation to conjecture that: $$\frac{p_n}{q_n} = 1 -\frac{1}{3} + \frac{1}{5} - \cdots + (-1)^n \frac{1}{2n+1} $$
for all $n \ge 0$. The first case is done, as stated above. Suppose the hypothesis for $n-1$ is true. Then \begin{align*}
\frac{p_n}{q_n} & = \frac{2 p_{n-1}}{q_n} + \frac{(2n-1)^2 p_{n-2}}{q_n} \\
& = 2\bigg(1-\frac{1}{3} + \cdots (-1)^{n-1}\frac{1}{2n-1}\bigg)\frac{q_{n-1}}{q_n} +(2n-1)^2 \bigg(1 - \frac{1}{3} + \cdots + (-1)^{n-2} \frac{1}{2n-3}\bigg) \frac{q_{n-2}}{q_n} \\
&=  \bigg(1 - \frac{1}{3} + \cdots + (-1)^{n-2} \frac{1}{2n-3}\bigg) + 2\frac{q_{n-1}}{q_n} (-1)^{n-1}\frac{1}{2n-1} \\
&= \bigg(1 - \frac{1}{3} + \cdots + (-1)^{n-2} \frac{1}{2n-3} +(-1)^{n-1}\frac{1}{2n-1}\bigg) + (2n-1)\frac{q_{n-2}}{q_n} (-1)^n \quad -\star
\end{align*}
Where I applied IH to both terms $p_{n-1}$ and $p_{n-2}$ in the second line. I now claim that $\frac{q_{n-2}}{q_n} =\frac{1}{(2n-1)(2n+1)}$ for all $n$. As usual, the base case is easily verified. Now suppose the hypothesis for $n-1$ is true. Then \begin{align*}
q_{n-2} &= 2q_{n-3} + (2n-5)^2 q_{n-4} \\
& = \frac{2q_{n-1}}{(2n-3)(2n-1)} + (2n-5)\frac{q_{n-2}}{2n-3} \quad \text{Apply IH to both terms} \\
& =\frac{1}{(2n-3)(2n-1)}\bigg(2 q_{n-1} + (2n-5)(2n-1) q_{n-2} \bigg) \\
&= \frac{1}{(2n-3)(2n-1)} \bigg(q_n - 4(2n-1)q_{n-2}\bigg)
\end{align*}
You can check that this easily simplifies to $\displaystyle \frac{q_{n-2}}{q_n}=\frac{1}{(2n-1)(2n+1)}$, which is what we want! Now substituting this back into the $(\star)$ we have proven by induction that:
$$\frac{p_n}{q_n} = 1 -\frac{1}{3} + \frac{1}{5} - \cdots + (-1)^n \frac{1}{2n+1} $$
So if we let $n\rightarrow \infty$, we see that this term is just equal to the series $$1-\frac{1}{3}+\frac{1}{5} - \frac{1}{7} + \cdots $$ This is precisely equal to the $\arctan 1$! (Search up the Taylor series expansion for arctan for this.) So this value is just $\displaystyle \frac{\pi}{4}$, which is what we desired.
