Investigate the convergence or divergence of the sequence a_{1}=\sqrt{c}, a_{2}=\sqrt{c+a_{1}}, a_{n+1}=\sqrt{c+a_{n}},



Answered question


Investigate the convergence or divergence of the sequence
I proved it is increasing. But I could not prove it is bounded above or unbounded. I guess it is divergence sequence. How can I do this?

Answer & Explanation



Beginner2022-01-16Added 33 answers

Claim: an is increasing and bounded above by
M=1+1+4c2. Also limnan=M
Proof: By induction on n1.
a2=c+c>c=a1. Assume an>an1, we show an+1>an.
But an+1=c+an>c+an1=an by the inductive step. So by induction an+1>an,n1. We show that an<M,n1 by induction also. For
If c<14, we're done since LHS<0<RHS.
Otherwise, c>142c1>0, and squaring both sides: 4c4c+1<1+4c4c<0 is clearly true. Thus a1<M. Assume an<M, you have an+1=c+an<Man<M2c
which is true by inductive step. So an<M,n1.
Thus the claim is verified. This follows that the limit exists and call it M. Then M=c+MM2=c+MM2Mc=0M=1±1+4c2
Since an>0,n1M0M=1+1+4c2
Wendy Boykin

Wendy Boykin

Beginner2022-01-17Added 35 answers

First, you need to study the fixed points of the function
as only these can be the limit of the sequence
Solving y=c+y, we find
The sequence starts at x=0 and is nonnegative, so the only possible candidate for the limit is
Now, we need to prove that if 0<x<y then x<f(x)<y.
Thus, the sequence is bounded from above by y and is increasing. As y is the only possible limit, it converges to y.

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