# Find the limit of the sequence

1. Apr 12, 2005

Find the limit of the sequence whose terms are given by $$a_n = (n^2)(1-cos(5.2/n))$$

well as n->inf, cos goes to 1 right? so shouldnt the limit of this sequence be 0?

2. Apr 12, 2005

### saltydog

Mathematica returns:

$$\mathop \lim\limits_{n\to \infty}n^2[1-Cos(\frac{a}{n})]=\frac{a^2}{2}$$

I'd like to know how too.

3. Apr 12, 2005

### shmoe

But $$n^2$$ goes to infinity, so your limit is an indeterminate form, infinity*0. Rewrite as:

$$a_n=\frac{1-\cos{\frac{5.2}{n}}}{\frac{1}{n^2}}$$

and use l'hopital, or you could use the taylor series for cos.

4. Apr 12, 2005

### saltydog

Right, just twice. Thanks.