1
$\begingroup$

I need to find:

$\lim_{n\to \infty} a_n =\lim_{n\to \infty} \frac{1}{n^2 +n} , for: \forall n \in \Bbb N \setminus \{ 0 \}$

By using the Sandwich a.k.a. squeeze theorem.My ideas so far:

when i factor out the n, i get:

$a_n = \frac{1}{n(n +1)}$

from that i can build the following inequations:

1: $0< \frac{1}{n} \le 1 , for: \forall n \in \Bbb N \setminus \{ 0 \}$

2: $0< \frac{1}{n+1} \le \frac{1}{2} , for: \forall n \in \Bbb N \setminus \{ 0 \}$

now when i multiply first inequality with $\frac{1}{n+1}$ i will get the following:

$0\left( \frac{1}{n+1} \right) < \frac{1}{n(n +1)} \le \left( \frac{1}{n+1} \right)1 $

or if i multiply the second inequality with $\frac1n$ i will get:

$0\left( \frac1n \right) < \frac{1}{n(n +1)} \le \left( \frac1n \right)1 $

My Problem with this approach:

The leftmost side of this inequality is smaller but not smaller or equal then the middle sequence, but i don't know if this is valid, and i dont know if one can simply say $\lim_{n \to \infty} 0 = 0$ either.

The second Problem is i would like to know if i choose the flanking sequences the right way.

  • 0
    There are about million ways (actually infinitely many ways) to choose the right `flanking' sequences. Yours is one of those.2012-11-08

1 Answers 1

1

$0< a_n <1/n$ for all $n$. Then you apply sandwich theorem.

  • 0
    To user1636457 : yes ...2012-11-08