Find the limit of the sequence defined by $\forall n\in \mathbb{N}^*,u_n = \frac{2n(2n-1)}{4n^2+1} u_{n-1}$ and $u_0 =1$

Question

Let $(u_n)$ be the sequence defined by $\forall n\in \mathbb{N}^*,u_n = \frac{2n(2n-1)}{4n^2+1} u_{n-1}$ and $u_0 =1$. I would like to find another way to find the limit of $(u_n)$.

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What I have done:

We can see that $(u_n)$ is a strictly positive decreasing sequence so it converges, and by induction we can show that :

$$\forall n \in \mathbb{N}, u_n = \displaystyle \prod_{k=1}^{n} \dfrac{2k(2k-1)}{4k^2+1}$$

And I end up with:

$$\forall n \in \mathbb{N}, u_n = \displaystyle \prod_{k=1}^{n} \dfrac{2k(2k-1)}{4k^2+1} \leq \prod_{k=1}^{n} \dfrac{2k(2k-1)}{4k^2} = \prod_{k=1}^{n} \dfrac{2k-1}{2k} $$

I managed to show that:

$$ \prod_{k=1}^{n} \dfrac{2k-1}{2k} = \dfrac{(2n)!}{(2^nn!)^2} \stackrel{\text{stirling}}{\sim} \dfrac{1}{\sqrt{n \pi}} $$

Therefore $\displaystyle \prod_{k=1}^{n} \dfrac{2k-1}{2k} \underset{n\rightarrow +\infty}{\longrightarrow} 0$ and by the squeeze theorem we deduce that $\displaystyle \lim_{n\rightarrow +\infty }u_n=0 $.

Do you see another approach?

Answer 1

I would proceed as follows: you already know that $u_n$ converges to a non negative limit. Also you can write $$u_n = \prod_{k=1}^{n}\frac{2k(2k-1)}{4k^2+1} = \prod_{k=1}^{n}\left(1-\frac{2k+1}{4k^2+1}\right)$$ By a well-known theorem this product is divergent, since $\sum_{k=1}^{\infty}\frac{2k+1}{4k^2+1}$ is divergent. Moreover, as $u_n$ converges being monotone decreasing and bounded below, it follows that this product must diverge to $0$, which is your limit.

This might be useful: Equivalence of convergence of a series and convergence of an infinite product Even though this link refers to positive $a_n$'s, it's still valid when all the $a_n$'s are negative.