PlanetMath: $(1 + 1/n)^n$ is an increasing sequence

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is an increasing sequence

(Theorem)

Theorem 1 The sequence is increasing.

Proof. To see this, rewrite

and divide two consecutive terms of the sequence:

$\displaystyle { \left( 1 + {1 \over n} \right)^n \over \left( 1 + {1 \over n-1} \right)^{n-1} }$	$\displaystyle =$	$\displaystyle { \left( {n+1 \over n} \right)^n \over \left( {n \over n-1} \right)^{n-1} }$
	$\displaystyle =$	$\displaystyle \left( {(n-1)(n+1) \over n^2} \right)^{n-1} {n+1 \over n}$
	$\displaystyle =$	$\displaystyle \left( 1 - {1 \over n^2} \right)^{n-1} \left( 1 + {1 \over n} \right)$

Since $(1 - x)^n \ge 1 - nx$ , we have

$\displaystyle { \left( 1 + {1 \over n} \right)^n \over \left( 1 + {1 \over n-1} \right)^{n-1} }$	$\displaystyle \ge$	$\displaystyle \left( 1 - {n-1 \over n^2} \right) \left( 1 + {1 \over n} \right)$
	$\displaystyle =$	$\displaystyle 1 + {1 \over n^3}$
	$\displaystyle >$	$\displaystyle 1,$

hence the sequence is increasing. $\qedsymbol$

Theorem 2 The sequence $(1 + 1/n)^{n+1}$ is decreasing.

Proof. By an inequality for differences of powers, we have

$\displaystyle (n^2)^n - (n^2 - 1)^n \ge n (n^2 - 1)^{n-1}.$

From this we may conclude

The last line follows from the factorization

. Dividing through,

$\displaystyle {n^n \over (n - 1)^n} > {(n + 1)^{n + 1} \over n^{n + 1}} ,$

which simplifies to

$\displaystyle \left( {n \over n - 1} \right)^n > \left( {n + 1 \over n} \right)^{n + 1},$

$\displaystyle \left( 1 + {1 \over n - 1} \right)^n > \left( 1 + {1 \over n} \right)^{n + 1}.$

$\qedsymbol$

is an increasing sequence" is owned by rspuzio.

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is an increasing sequence, born on 2006-03-25, modified 2007-05-16.
Object id is 7775, canonical name is 11nnIsAnIncreasingSequence.
Accessed 6649 times total.

Classification:

33B99 (Special functions :: Elementary classical functions :: Miscellaneous)

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