Question: How do we show that the sequence (1+(1/n))^n is bounded?

Posted by: Ashley on 20.10.2020

We 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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