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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?

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  • $\begingroup$ 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. $\endgroup$ – fractal1729 May 21 '17 at 22:56
  • $\begingroup$ Would this now be correct? $\endgroup$ – user6190474 May 22 '17 at 5:24
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hint

First, you need to prove that both sequences are convergent.

for this, it is easy to show by induction that

$$ k_n\leq k_{n+1}\leq t_{n+1}\leq t_n$$

which means that $(k_n)$ and $(t_n) $ are monotonic bounded.

then use your method.

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