Show that

Exercise n 0.

Show that

 1 − _n ¹

 n

is bounded

0 ≤

 1 − 1

n

 ⁿ

< 1 Exercise n.1

Show that the sequence

x n =

 1 − 1

n

 ⁿ , is an increasing sequence. We have

x ₂ > x ₁ For n > 1 consider the ratio

x n+1

x n

=



1 − _n+1 ¹

 n+1

 1 − _n ¹

 n =

 n n + 1

 n+1  n n − 1

 n

=

 n n + 1

 ⁿ⁺¹  n n − 1

 ⁿ⁺¹  n − 1 n



=

 n

(n + 1) n (n − 1)

 ⁿ⁺¹  n − 1 n



=

 n ² (n ² − 1)

 n+1

 n − 1 n



=  n ² − 1 + 1 (n ² − 1)

 n+1

 n − 1 n



=  n ² − 1 n ² − 1 + 1

n ² − 1

 n+1  n n − 1



By Bernoulli inequality with

0 < h = 1

n ² − 1 , ∀n > 1 Hence, substituing in the previous inequality

x _n+1 x n

=

 1 + 1

n − 1

 n n − 1



> 1, ∀n > 1 Also, the sequence (x n ) defined by

x n =  1 − 1

n

 ⁿ

, n = 1, 2, . . . .

is increasing by an application of inequality between geometrical and arithmetic mean. Indeed, applying the inequality

n+1

√

a 1 . . . a n+1 ≤ a ₁ + · · · + a _n+1 n + 1 with

a ₁ = · · · = a _n = 1 − 1

n and a _n+1 = 1, we obtain

n+1

r

 1 − 1

n

 ⁿ

≤ n

n + 1 = 1 − 1 n + 1 . This is equivalent to x n ≤ x n+1 . Hence (x n ) is increasing.

Exercise n 2. Compute

n→∞ lim

 1 − 1

n ²

 ⁿ .

1

2

By Bernoulli inequality

1 − 1 n <

 1 − 1

n ²

 n

< 1 Hence, passing to the limit

n→∞ lim

 1 − 1

n ²

 ⁿ

= 1 Exercise n 3.

From the previous exercizes compute

n→∞ lim

 1 − 1

n

 ⁿ . We have

n→∞ lim

 1 − 1

n

 ⁿ

= 1 e

n→∞ lim

 1 − 1

n ²

 n

= lim

n→∞

 1 − 1

n

 n n→∞ lim

 1 + 1

n

 n

then

n→∞ lim

 1 − 1

n

 n

=

lim n→∞

 1 − _n ¹

 n

lim n→∞

 1 + ¹ _n

 n = 1 e

References

[1] E. Giusti, Analisi Matematica I, Boringhieri Ed, 1988.

Dipartimento di Metodi e Modelli, Matematici per le Scienze Applicate, Universit` a di Roma “La Sapienza”, Via A. Scarpa, 16, 00161 Roma, Italy

E-mail address: loreti@dmmm.uniroma1.it