For the statement "convergence implies boundedness," the converse statement would be "boundedness implies convergence."
So, we are asking if "boundedness implies convergence" is a true statement.
Pf//
By way of contradiction, "boundedness implies convergence" is false.
Let the sequence (Xn) be defined as
Xn = 1 if n is even and
Xn = 0 if n is odd.
So, (Xn) = {X1,X2,X3,X4,X5,X6...} = {0,1,0,1,0,1,...}
Note that this is a divergent sequence.
Also note that for all n, -1 < Xn < 2
Therefore, the sequence (Xn) is bounded above by 2 and below by -1.
As we can see, we have a bounded function that is divergent. Therefore, by way of contradiction, we have proven the converse false.
Q.E.D.
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That's true. If a function is continuous, it's (Riemman) integrable, but the converse is not true.
No
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