Posted by: Ashley on 20.10.2020
Ask a QuestionWe can use limits for this proof.
Let y=(1+(1/n))^n
Taking "ln()" on both sides:
ln(y) = ln((1+(1/n))^n) = n ln(1+(1/n)) = [ln(1+(1/n))] / [1/n]
Let x=1/n. Then, as n tends to infinity, x tends to 0. Also,
ln(y) = ln(1+x) / x
Taking limit n tends to infinity, or x tends to 0 on both sides, we have
lim(n->infinity) ln(y) = lim(x->0) (ln(1+x) / x)
=> lim(n->infinity) ln(y) = lim(x->0) (ln(1+x) / x)
The limit on the RHS is in 0/0 form Let us apply L'Hospital's rule and differentiate the numerator and denominator with respect to x. Then,
lim(n->infinity) ln(y) = lim(x->0) ( (1/(1+x)) / 1)= lim(x->0) 1/(1+x) = 1
=> lim(n->infinity) ln(y) = 1
Raising both sides to the power of e, we have
lim(n->infinity) e^(ln(y))=e^1
=> lim(n->infinity) y = e
=> lim(n->infinity) (1+(1/n))^n = e
So, the given sequence tends to e as n tends to infinity, and we know that:
Every convergent sequence of real numbers must be bounded.
So, the sequence (1+(1/n))^n is bounded.
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