Solution

First suppose

x_{n} \to L > 0

. Continuity of the natural logarithm gives

lo g x_{n} \to lo g L

. Part (a), applied to this sequence, yields

\frac{1}{n} \sum_{k = 1}^{n} lo g x_{k} ⟶ lo g L .

Applying the continuous exponential function,

n x_{1} x_{2} \dots x_{n} = exp (\frac{1}{n} \sum_{k = 1}^{n} lo g x_{k}) ⟶ L .

L = 0

, then for every

ε > 0

all sufficiently late

x_{k}

are below

ε

. Splitting off the fixed initial product shows that the limsup of the geometric means is at most

ε

; hence it is zero. This proves the geometric mean of a sequence result in all cases.

Now suppose

r_{n} = \frac{x _{n}}{x _{n - 1}} ⟶ r .

The telescoping product is

x_{n} = x_{0} \prod_{k = 1}^{n} r_{k} .

Therefore

n x_{n} = x_{0}^{1/ n} n r_{1} r_{2} \dots r_{n} ⟶ 1 \cdot r,

and so

n \to \infty lim n x_{n} = n \to \infty lim \frac{x _{n}}{x _{n - 1}} .

Solved by gpt-5.6-sol high.

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