Where does this sequence converge?
√7,√7+√7,√7+√7+√7,...
Answer
For a proof of convergence,
Define the sequence as
x0=0
xn+1=√7+xn
Note that xn≥0 ∀n.
Notice that x2−x−7=(x−a)(x−b) where a<0 and b>0.
We claim the following:
i) xn<b⟹xn+1<b
ii) xn<b⟹xn+1>xn
For a proof of i)
We have that
xn<b=b2−7 and so xn+7<b2 and thus by taking square roots xn+1<b
For a proof of ii)
We have that
(xn+1)2−(xn)2=−(x2n−xn−7)=−(xn−a)(xn−b)>0 if xn<b.
Thus {xn} is monotonically increasing and bounded above and so is convergent.
By setting L=√7+L, we can easily see that the limit is b=1+√292
In fact, we can show that the convergence is linear.
b−xn+1b−xn=b2−(7+xn)(b+√7+xn)(b−xn)=1b+xn+1
Thus lim.
We can also show something a bit stronger:
Let \displaystyle t_n = b - x_n.
The we have shown above that \displaystyle t_n \gt 0 and \displaystyle t_n \lt b^2
We have that
\displaystyle b - t_{n+1} = \sqrt{7 + b - t_n} = \sqrt{b^2 - t_n}
Dividing by \displaystyle b throughout we get
\displaystyle 1 - \dfrac{t_{n+1}}{b} = \sqrt{1 - \dfrac{t_n}{b^2}}
Using \displaystyle 1 - \dfrac{x}{2} \gt \sqrt{1-x} \gt 1 - x \ \ 0 \lt x \lt 1 we have that
\displaystyle 1 - \dfrac{t_n}{2b^2} \geq 1 - \dfrac{t_{n+1}}{b} \geq 1 - \dfrac{t_n}{b^2}
And so
\displaystyle \dfrac{t_n}{2b} \leq t_{n+1} \leq \dfrac{t_n}{b}
This gives us that \displaystyle b - \dfrac{b}{b^n} \leq x_n \leq b - \dfrac{b}{(2b)^n}
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