I am having an issue understanding why the sequence $x_n$ such that $x_{n+1} = \sqrt{2x_n +1}$, $x_1 = 1$ is convergent. After plugging numbers into the sequence we discovered that the sequence is increasing and we could not find a limit that would make it bounded and therefore not convergent. Any ideas?
Asked
Active
Viewed 61 times
1
-
A bounded monotonic function is convergent, it may be increasing but if it has a upper bound then it is convergent. You can find the upper bound from induction – John Dec 15 '13 at 18:08
2 Answers
1
Hint: You can prove by induction that (for example) $x_n\lt 4$ for all $n$.
The numerical evidence that the sequence is increasing is useful. But a proof should be given (again by induction) that the sequence is increasing.
André Nicolas
- 507,029
-
I understand that the sequence is increasing but how do i show that it is bounded? – user115964 Dec 15 '13 at 18:55
-
-
Induction step: Suppose that $a_k\lt 4$. Then $a_{k+1}=\sqrt{2a_k+1}\lt \sqrt{(2)(4)+1}=3\lt 4$. – André Nicolas Dec 15 '13 at 18:57
0
You must use mathematical induction to prove that $a_{n+1}>a_{n}$. It is obviously true for $n=1$ So, I will jump to the main part: If we assume that its also true for $n=k \implies a_{k+1}>a_{k}$ Now, we have to prove that is true for $n=k+1$
If we assume, $$a_{k+1}>a_{k}$$ $$2a_{k+1}+1>2a_{k}+1$$ $$\sqrt{2a_{k+1}+1}>\sqrt{2a_{k}+1}$$ $$\implies a_{k+2}>a_{k+1}$$ Therefore, since for the case $n=1$ the proposition is true and whenever $n=k$ holds, $n=k+1$ is true, then $a_{n+1}>a_{n}$ by PMI. This is basically saying that the sequence is increasing.
But, we know that $$a_{k+1}>a_{k}$$ $$\implies a_{k}<a_{k+1}=\sqrt{2a_{k}+1}$$ $$\iff a^2_{k}<2a_k+1$$ $$a^2_{k}-2a_k+1<2$$ $$(a_k-1)^2<2$$ $$\implies a_k<\sqrt{2}+1$$
Hence, our sequence is upper bounded and increasing, therefore by the monotonic convergence theorem the sequence is convergent.
Notice that we could have done it the way Andre Nicholas suggested but if the question asked for the limit of the sequence, do not mistake the upper-bound for the limit. In my case, the upper-bound is the limit. This can be proven as such:
$$\text{Assume} \lim_{n \to \infty} a_n = L \text{}$$ We can only do this because we proved earlier that there exists a limit. You can't assume there is a limit and find the limit and therefore say there is a limit. $$\lim_{n \to \infty} a_{n+1}=\lim_{n \to \infty} \sqrt{2a_n+1}=\sqrt{2L+1}$$ But, we know that $$\lim_{n \to \infty} a_n=\lim_{n \to \infty} a_{n+1}=L$$ $$\implies \sqrt{2L+1}=L$$ The solution for L is $\sqrt{2}+1$ not $\sqrt{2}-1$ because it is an increasing sequence.
John
- 1,149