# Prove that $(k_n)$ and $(t_n)$ converge to the same limit.

Let $0 < k_1 < t_1$ and for every integer $n \geq 1$, set $k_{n+1}= \sqrt{k_nt_n}$ and $t_{n+1}=(k_n+t_n)/2$. Prove that $(k_n)$ and $(t_n)$ converge to the same limit.

This is what I had before...

Let $0 < k_1 < t_1$ and $k_{n+1}= \sqrt{k_nt_n}$, $t_{n+1} =(k_n+t_n)/2$, $\forall n \in \mathbb{N}$ be given. Since $0 \leq t_{n+1}-k_{n+1} \leq (t_1-k_1)/2^n$, $\forall n \in N$ then by the squeeze theorem, lim$_{n \rightarrow \infty} (t_{n+1}-k_{n+1})= 0$. Therefore, lim$_{n \rightarrow \infty} t_n =$ lim$_{n \rightarrow \infty} k_n$.

Now with some improvments...

Let $0 < k_1 < t_1$ and $k_{n+1}= \sqrt{k_nt_n}$, $t_{n+1} =(k_n+t_n)/2$, $\forall n \in \mathbb{N}$ be given. Using the arithmetic-geometric mean inequality, then $k_{n+1}= \sqrt{k_nt_n} \leq (k_n+t_n)/2 = t_{n+1}$, $\forall n \in \mathbb{N}$. Thus, $0 < k_1 \leq t_1$, $\forall n \in \mathbb{N}$. Since $t_{n+1} =(k_n+t_n)/2$ then $t_{n+1} =(k_n+t_n)/2 \leq (t_n+t_n)/2 = t_n$. Similarly, since $k_{n+1}= \sqrt{k_nt_n}$ then $k_{n+1}= \sqrt{k_nt_n} \geq \sqrt{k_nk_n} = \sqrt{k_n^2}= k_n$. Therefore, $K_n \leq t_n$, $\forall n \in \mathbb{N}$ and hence $k_n \leq k_{n+1}\leq t_{n+1}\leq t_n$. Since $k_n \leq k_{n+1}\leq t_{n+1}\leq t_n$ then $0 \leq t_{n+1}-k_{n+1} \leq (t_1-k_1)/2^n$, $\forall n \in N$ then by the squeeze theorem, lim$_{n \rightarrow \infty} (t_{n+1}-k_{n+1})= 0$. Therefore, lim$_{n \rightarrow \infty} t_n =$ lim$_{n \rightarrow \infty} k_n$. .

Is this how you would correctly prove this/is it written well?

• You will have to elaborate on how you obtained $0 \leq t_{n+1} - k_{n+1} \leq (t_1 - k_1)/2^n$; that is a nontrivial claim. – fractal1729 May 21 '17 at 22:56
• Would this now be correct? – user6190474 May 22 '17 at 5:24

$$k_n\leq k_{n+1}\leq t_{n+1}\leq t_n$$
which means that $(k_n)$ and $(t_n)$ are monotonic bounded.