# limit of $u_n = u_{n-1}\cdot \frac{n}{n+1}$ and $v_n = v_{n-1}\cdot \frac{n+1}{n}$

What are the limits of $u_n = u_{n-1}.\frac{n}{n+1}$ and $v_n = v_{n-1}.\frac{n+1}{n}$ ?

Seems that $u_n \to 0$ and $v_n \to \infty$, but I can't prove it.

1. For $u_n$ we have an infinite product of term all strictly lower than $1$. Can we say that goes to 0, and why ?
2. For $v_n$ we have an infinite product of teram all strictly greater than $1$. Can we say that goes to infinity, and why ?

Edit : by induction, $u_n = \frac{u_{1}}{n+1}$ and $v_n = v_{1}.({n+1})$. and thus the result.

But does 1. and 2. are valid statements (Maybe I need to open an other question?).

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Regarding your for u/v questions: Both assumptions are false in general, common counter-examples are can be found on en.wikipedia.org/wiki/Euler_product: $\prod\limits_p (1-p^{-2})^{-1}=\zeta(2)$ has factors strictly $>1$ and $\prod\limits_p (1+p^{-2})^{-1}=\zeta(4)/\zeta(2)$ has factors strictly $<1$. –  gammatester Oct 28 '13 at 13:01
There is absolutely no need to consider these as infinite products and, in my opinion, it only clouds the issue by introducing a lot of unnecessary machinery –  TBrendle Oct 28 '13 at 13:02

Hint: Just write out $u(n)$ and $v(n)$ for $n \le 5$ or $n \le 10$, say.

Prove your conjectures for closed forms for $u(n)$ and $v(n)$ by induction.

Now the limits should be easy.

For your conjectures, consider the following identities:

$$1 = \sin\frac{\pi}2 = \frac\pi2\prod_{n=1}^\infty \left(1-\frac1{4n^2}\right)\qquad \sinh \frac\pi2 = \frac\pi2\prod_{n=1}^\infty\left(1+\frac1{4n^2}\right)$$

which demonstrate that such infinite products may converge to nontrivial values (for which a necessary condition is that the individual terms converge to $1$).

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Question edited for induction –  Timmy Oct 28 '13 at 12:53
@Timmy Yes, that's correct. I've added a bit on infinite products. –  Lord_Farin Oct 28 '13 at 12:55

The question is not well posed, since you didn't specify initial points. Fortunately, convergence of the first sequence doesn't depend on its initial point.

You have $$u_n = u_{n-1}\frac{n}{n+1}.$$ Then also $$u_{n+1} = u_{n} \frac{n+1}{n+2}.$$ Combining these facts, we have $$u_{n+1} = u_{n-1}\frac{n}{n+2}.$$ Show by induction that $$u_{n+k} = u_{n-1} \frac{n}{n+k+1}.$$ Then since $$\lim_{k \to \infty} \frac{n}{n+k+1}=0,$$ we have $u_{n+k} \to 0$ as $k \to \infty$. This implies $u_n \to 0$ as $n \to \infty$.

The behavior of the other sequence will depend on the initial point $v_0$. If $v_0<0$ then it diverges to $-\infty$, if positive then to positive infinity, and if $v_0=0$ then the sequence converges trivially to 0. Since you didn't specify an initial point, just a recurrence relation, we have no way to decide.

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Fix $u_0$ and $v_0$ as any arbitrary numbers. The idea that follows is independent of the values ​​that $u_0$ and $v_0$ assume. $$\begin{array}{rcl} u_n&=& u_{0}\frac{1}{1+1}\cdot\frac{2}{2+1}\cdot\cdot\frac{3}{3+1}\cdot \ldots\cdot\frac{(n-1)}{(n-1)+1} \cdot\frac{n}{n+1}\\ u_n&=& u_{0}\frac{1}{2}\cdot\frac{2}{3}\cdot\cdot\frac{3}{4}\cdot \ldots\cdot\frac{(n-2)}{n-1}\cdot\frac{(n-1)}{n} \cdot\frac{n}{n+1}\\ &=& u_{0}\frac{1}{n+1}\longrightarrow 0\\ \end{array}$$ So we entirely analogous $$v_n=v_0\cdot \frac{2}{1}\cdot\frac{3}{2}\cdot\cdot\frac{4}{2}\cdot \ldots\cdot\frac{(n-1)}{n-2}\cdot\frac{(n-2)}{n-1} \cdot\frac{n-1}{n+1}\cdot \cdot\frac{n}{n+1}\\ =v_0\cdot(n+1)\longrightarrow \mbox{sgn}(u_0)\infty, \quad\mbox{ if } v_0\neq 0.$$ In general if the product is a telescopic product ( exemple $\prod_{k=1}^n\frac{f(k)}{f(k-1)}=\frac{f(n)}{f(0)}$ ) and there is $\lim_{k\to \infty}\frac{f(n)}{f(0)}=L\in\mathbb{R}\cup\{-\infty,+\infty\}$ then $$x_n= \begin{cases} x_0,& \mbox{ if } n=0\\ \prod_{k=1}^n\frac{f(k)}{f(k-1)}x_0,& \mbox{ if } n>0\\ \end{cases} \implies \lim_{n\to\infty}x_n=L\cdot x_0$$

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