# Finding lower bound sequence for squeeze theorem

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 inequation 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 inequation 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 inequation is smaller but not smaller or equal then the middle sequence, but i dont now 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.

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smaller is in particular smaller or equal. It is trivial to verify that $\lim_{n\to \infty} 0=0$ by the definition of limit. – Lior B-S Nov 8 '12 at 11:19
There are about million ways (actually infinitely many ways) to choose the right `flanking' sequences. Yours is one of those. – Lior B-S Nov 8 '12 at 11:21
Thank you for your help! – user1636457 Nov 8 '12 at 11:29

$0< a_n <1/n$ for all $n$. Then you apply sandwich theorem.
Do you mean that after if made $0<a_n< \frac 1n , \forall n \in \Bbb N \setminus \{ 0 \}$ i can just compute the $\lim_{n \to \infty} 0 = \lim_{n \to \infty} \frac 1n = 0 \implies lim_{n \to \infty} a_n = 0$ ? – user1636457 Nov 8 '12 at 16:29