# 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 $u_0 =1$

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 $$(u_n)$$.

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?

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.

• Does your well-known theorem has a name? Could you provide more details? Well I disagree, if $a_n = -1$ for all $n$ then the result is false. Thank you for the effort but your answer is not clear enough for me.
– Axel
Dec 10, 2020 at 17:58
• When $a_n = -1$ for all $n$ the product is not convergent in the product sense. See here for more details for example en.wikipedia.org/wiki/…. Dec 10, 2020 at 18:04
• A necessary condition for the convergence of any product $\prod_{k=1}^{\infty}(1+a_k)$ is certainly $a_k \to 0$ as $k \to +\infty$ Dec 10, 2020 at 18:05
• You should learn about them! There's a very nice theory which relates them to infinite series. Dec 10, 2020 at 18:07
• I realized afterwards that I did not need any help. But thank you for introducing me this notion of infinite product.
– Axel
Dec 11, 2020 at 15:30