Investigate the convergence or divergence of the sequence a1=c,a2=c+a1,an+1=c+an,c&gt;0. 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?

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Investigate the convergence or divergence of the sequence  a1=c,a2=c+a1,an+1=c+an,c&amp;gt;0.  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?

Medicim6 · Accepted Answer

Claim: an is increasing and bounded above byM=1+1+4c2. Also limn→∞an=MProof: By induction on n≥1.a2=c+c&amp;gt;c=a1. Assume an&amp;gt;an−1, we show an+1&amp;gt;an.But an+1=c+an&amp;gt;c+an−1=an by the inductive step. So by induction an+1&amp;gt;an,∀n≥1. We show that an&amp;lt;M,∀n≥1 by induction also. Forn=1,a1=c&amp;lt;M⇔c&amp;lt;1+1+4c2⇔2c−1&amp;lt;1+4cIf c&amp;lt;14, we&#039;re done since LHS&amp;lt;0&amp;lt;RHS.Otherwise, c&amp;gt;14⇒2c−1&amp;gt;0, and squaring both sides: 4c−4c+1&amp;lt;1+4c⇔−4c&amp;lt;0 is clearly true. Thus a1&amp;lt;M. Assume an&amp;lt;M, you have an+1=c+an&amp;lt;M⇔an&amp;lt;M2−c=1+21+4c+1+4c4−c=1+1+4c2=Mwhich is true by inductive step. So an&amp;lt;M,∀n≥1.Thus the claim is verified. This follows that the limit exists and call it M. Then M=c+M⇒M2=c+M⇒M2−M−c=0⇒M=1±1+4c2Since an&amp;gt;0,∀n≥1⇒M≥0⇒M=1+1+4c2

Wendy Boykin · Accepted Answer

First, you need to study the fixed points of the function  f(x)=c+x  as only these can be the limit of the sequence  f(0),f(f(0)),f(f(f(0))),…  Solving y=c+y, we find  y±=1±1+4c2  The sequence starts at x=0 and is nonnegative, so the only possible candidate for the limit is  y=1+1+4c2  Now, we need to prove that if 0&amp;lt;x&amp;lt;y then x&amp;lt;f(x)&amp;lt;y.  x&amp;lt;f(x)=c+x&amp;lt;c+y=f(y)=y  Thus, the sequence is bounded from above by y and is increasing. As y is the only possible limit, it converges to y.

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}},

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