Let $x_{n+1}^2=x_n+\frac{1}{n+1}$, $x_{n+1}>0, x_1=1$. How to prove that $x_n\to 1$?

Question

Let $x_{n+1}^2=x_n+\frac{1}{n+1}$, $x_{n+1}>0, x_1=1$. How to prove that $x_n\to 1$?

$x_{n+1}<x_n+\frac{1}{n+1}$ seems not enough!

Any ideas? Compuationally, $n$ large $x_n$ decresing. How to prove mathematically?

Answer 1

To prove $x_n \to 1$, use the following steps:

Mathematical Induction

1. Base case: $x_1 = 1$ (given).
2. Inductive step: Assume $x_k = 1$ for some $k$. Prove $x_{k+1} = 1$.

Sequence Definition Given $x_{2n+1} = x_n \cdot x_{n+1}$, we can rewrite it as:

Recurrence Relation $x_{2n+1} = x_n \cdot x_{n+1}$

Convergence Proof

1. Since $x_{n+1} > 0$, the sequence is positive.
2. Show the sequence is bounded: $1 \leq x_n$ for all $n$.
3. Use the recurrence relation to prove $x_{2n} \leq x_{2n+1}$.
4. Apply the Monotone Convergence Theorem.

Limit Proof

1. Let $L = \lim_{n \to \infty} x_n$.
2. From the recurrence relation: $L = L^2$.
3. Solve for $L$: $L = 1$.

Detailed Solution For a more detailed solution, consider the following:

1. $x_1 = 1$.
2. $x_3 = x_1 \cdot x_2 = 1 \cdot x_2$.
3. $x_5 = x_2 \cdot x_3 = x_2 \cdot (1 \cdot x_2) = x_2^2$.
4. $x_{2n+1} = x_n \cdot x_{n+1} = x_{n-1}^{2^{n-1}} \cdot x_n^{2^{n-1}}$.

This pattern suggests $x_n$ converges to 1.