Setting an=√2+√2+√2+ I get that an+1=√2+ana1=√2. Clearly all numbers in the sequence are positive and we see that $a_n
We can use the help-function f(x)=√2+x such that an+1=f(an). But since f′(x)=12√2+x=0⇔no real solutions,
f(x) never flattens out or decreases, so it can't be convergent?
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