Let c epsilon R with 0 SolutionConsider cn and cn1 cn cn1

Let c epsilon R with 0

Consider cⁿ and c^n-1

cⁿ = c^n-1 * c and since 0 < c < 1, c^n-1 > cⁿ

Using the theorem in Hint 1, we can wrtie the above inequality as

1 + nc <= 1 + (n-1)c or 1 + nc <= 1 + nc - c

or c <= 0. But this becomes a contradiction unless c is a 0 or positive. This implies that cⁿ -> 0 thus proving that b increasing n, we can make c^n small. In fact the limit when n tends to infinity is 0.

Let c epsilon R with 0 SolutionConsider cn and cn-1 cn = cn-1 * c and since 0 < c < 1, cn-1 > cn Using the theorem in Hint 1, we can wrtie the above i

Let c epsilon R with 0 SolutionConsider cn and cn1 cn cn1

Solution

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