Show that $u_n$ converges if $\min(u_n,u_{n+1})$ converges. Let $\lambda \in (0,1)$ and $(u_n)_n$ a sequence of real numbers such that :  $\forall n \in \mathbb N : \; u_{n+2}\ge \lambda u_{n+1} + (1-\lambda)u_n$.
And $\forall n \in \Bbb N \; v_n=\min(u_{n+1},u_n)$ .
The problem first asks to show that $v_n$ does have a limit, I have done this by showing that it's an increasing sequence. This immediately gives us that if $v_n \to +\infty$ then $u_n \to +\infty$ .
The next question is to show that if $v_n \to l \in \Bbb R$ then $u_n$ converges. I fail to see why this must be true, I tried to use the definition of limits but to no avail.
 A: I don't see a straightforward way to prove this without using
contradiction.
Suppose $v_n \uparrow l$. Since $u_n \ge v_n$ we have $\liminf_n u_n \ge l$
and since $u_n = v_n$ infinitely often we have, in fact, $\liminf_n u_n = l$.
Suppose $u = \limsup_n u_n > l$.
In particular, there is some $\epsilon>0$ such that
$u_n \ge l+ \epsilon$ infinitely often.
Let $\delta >0$ and choose $N$ such that for $n \ge N$ we have
$v_n \ge l-\delta$.
Suppose $u_n \ge l+ \epsilon$ for some $n \ge N+1$, then we must have
$l \ge v_{n-1} = u_{n-1}$ and $l \ge v_n = u_{n+1}$.
Since $u_{n+1} \ge \lambda u_n + (1-\lambda) u_{n-1}$, this gives
$l \ge \lambda(l+\epsilon) +(1-\lambda)(l-\delta) = l + \lambda \epsilon - (1-\lambda) \delta$.
Since $\delta>0$ was arbitrary, we have a contradiction. Hence
$u=l$.
A: Here is a different style of proof that removes the need for $\delta$ and $\epsilon$.  [Ultimately it will be the same idea as the other proof]. 

Suppose $v_n \nearrow L$, where $L \in \mathbb{R}$. 
We know $u_n \geq v_n$ and so $\liminf_{n\rightarrow\infty} u_n \geq \liminf_{n\rightarrow\infty} v_n = L$.  
We want to show $\limsup_{n\rightarrow\infty} u_n \leq L$.  If $u_n>L$ only for a finite number of indices $n$, we are done.  Else, let $\{u_{n[k]}\}_{k=1}^{\infty}$ denote the infinite subseqeunce of times such that $u_{n[k]}>L$. So $\lim_{k\rightarrow\infty} n[k]=\infty$. We know $v_i \leq L$ for all $i$, and so for each $k$ such that $n[k]-1\geq 1$, it must be that $v_{n[k]-1}=u_{n[k]-1}$ and $v_{n[k]}=u_{n[k]+1}$. So for all $k$ such that $n[k]-1\geq 1$: 
\begin{align}
L &\geq v_{n[k]}\\
&=u_{n[k]+1} \\
&\geq \lambda u_{n[k]} + (1-\lambda) u_{n[k]-1}\\
&= \lambda u_{n[k]} + (1-\lambda)v_{n[k]-1} 
\end{align}
Taking a $\limsup$ as $k\rightarrow\infty$ implies: 
$$ L \geq \lambda \limsup_{k\rightarrow\infty} u_{n[k]} + (1-\lambda) L $$
Since $\lambda>0$, this means $\limsup_{k\rightarrow\infty} u_{n[k]} \leq L$.
Since $u_n\leq L$ if $n$ is not in the subsequence $\{n[k]\}_{k=1}^{\infty}$, we know $\limsup_{n\rightarrow\infty} u_n \leq L$.

The rate of convergence can be arbitrarily slow: Fix $L>0$.  Let $f:\mathbb{N} \rightarrow \mathbb{R}$ be a function such that $f(1)=0$ and $f(n)$ increases arbitrarily slowly to $L$. Consider: 
\begin{align}
u_1 &=0\\
u_2 &= L-f(2)\\
u_3 &= \lambda u_2 + (1-\lambda) u_1 \\
u_4 &= L-f(4) \\
u_5 &= \lambda u_4 + (1-\lambda) u_3\\
u_6 &= L-f(6)
\end{align}
and so on, so 
\begin{align}
u_{n+2} &= \lambda u_{n+1} + (1-\lambda) u_n \quad, \mbox{ if $n+2$ is odd}\\
u_{n+2} &= L - f(n+2) \quad, \mbox{ if $n+2$ is even}
\end{align}
This satisfies the required inequality $u_{n+2} \geq \lambda u_{n+1} + (1-\lambda) u_n$ with equality when $n+2$ is odd, and with inequality when $n+2$ is even (since it defines $u_{n+2}$ as larger than all previous sequence values, and hence larger than the average of the past two sequence values).
Then $u_n\rightarrow L$, but arbitrarily slowly (according to the rate of convergence of $f$).
A: By the constraint on the sequence,
$$
\begin{align}
u_n
&\ge\lambda u_{n-1}+(1-\lambda)u_{n-2}\\[6pt]
&=\left\{\begin{array}{}
u_{n-1}+(1-\lambda)(u_{n-2}-u_{n-1})&&&&&\text{if }\min(u_{n-1},u_{n-2})=u_{n-1}\\
u_{n-2}+\lambda(u_{n-1}-u_{n-2})&&&&&\text{if }\min(u_{n-1},u_{n-2})=u_{n-2}
\end{array}\right.\\
&=\left\{\begin{array}{}
\min(u_{n-1},u_{n-2})+(1-\lambda)|u_{n-1}-u_{n-2}|&\text{if }\min(u_{n-1},u_{n-2})=u_{n-1}\\
\min(u_{n-1},u_{n-2})+\lambda|u_{n-1}-u_{n-2}|&\text{if }\min(u_{n-1},u_{n-2})=u_{n-2}
\end{array}\right.\\[6pt]
&\ge\min(u_{n-1},u_{n-2})\tag1
\end{align}
$$
since $u_{n-1}\ge\min(u_{n-1},u_{n-2})$, we get
$$
\min(u_n,u_{n-1})\ge\min(u_{n-1},u_{n-2})\tag2
$$
Thus, $v_n=\min(u_{n+1},u_n)$ is increasing. Inequality $(1)$ can be restated as
$$
u_n\ge v_{n-2}\tag3
$$
Case 1: $\boldsymbol{v_n}$ is unbounded.
Inequality $(3)$ says that 
$$
\begin{align}
\liminf\limits_{n\to\infty}u_n
&\ge\liminf\limits_{n\to\infty}v_{n-2}\\
&=\lim\limits_{n\to\infty}v_{n-2}\\
&=\infty\tag4
\end{align}
$$
Thus, $\lim\limits_{n\to\infty}u_n=\infty$.
Case 2: $\boldsymbol{v_n}$ is bounded above.
Since $v_n$ is increasing, $V=\lim\limits_{n\to\infty}v_n$ exists and is finite.
Inequality $(3)$ says that 
$$
\begin{align}
\liminf\limits_{n\to\infty}u_n
&\ge\liminf\limits_{n\to\infty}v_{n-2}\\
&=\lim\limits_{n\to\infty}v_{n-2}\\
&=V\tag5
\end{align}
$$
Suppose that $\boldsymbol{\limsup\limits_{n\to\infty}u_n\gt V}$. This means that there is an $\epsilon\gt0$ so that for all $N$, there is an $n\ge N$ so that $u_n\gt V+\epsilon$.
Find an $N$ so that if $n\ge N$ then $|v_n-V|\le\frac{\lambda\epsilon}4$. Find an $n\ge N+1$ so that $u_n\gt V+\epsilon$.
Since $v_{n-1},v_n\le V+\frac{\lambda\epsilon}4$, we know that $u_{n+1}=v_n$ and $u_{n-1}=v_{n-1}$.
Therefore,
$$
\frac{\lambda\epsilon}2\ge v_{n+1}-v_n=u_{n+1}-u_{n-1}\ge\lambda(u_n-u_{n-1})\ge\lambda\left(\epsilon-\frac{\lambda\epsilon}4\right)\ge\frac{3\lambda\epsilon}4\tag6
$$
Inequality $(6)$ implies our supposition is false; therefore, we must have
$$
\limsup_{n\to\infty}u_n\le V\tag7
$$
Inequalities $(5)$ and $(7)$ imply that
$$
\lim_{n\to\infty}u_n=V\tag8
$$
