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 limn→∞b−xn+1b−xn=12b.
We can also show something a bit stronger:
Let tn=b−xn.
The we have shown above that tn>0 and tn<b2
We have that
b−tn+1=√7+b−tn=√b2−tn
Dividing by b throughout we get
1−tn+1b=√1−tnb2
Using 1−x2>√1−x>1−x 0<x<1 we have that
1−tn2b2≥1−tn+1b≥1−tnb2
And so
tn2b≤tn+1≤tnb
This gives us that b−bbn≤xn≤b−b(2b)n
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