Bound of sequence $$a_n=\frac{n\cos(n)+\sin(n^2+2n-5)}{n^2}$$
I am trying to find the bound of this sequence.But I get stuck after some computations. I take $\sin f(x)\leq|f(x)| $ and $|\cos(x)|\leq1$ but I am not sure what to do at the end.Wasted 10 papers.
 A: When $n$ gets large so does $n^2+2n-5$. Since we don't have much control over $\cos x$ and $\sin x$ when $x$ is large we have to face the possibility that $\cos n$ and $\sin(n^2+2n-5)$ both are near $\pm1$ at the same time for certain large $n$. It follows that the best "universal" estimate is
$$|a_n|\leq{n+1\over n^2}\qquad(n\geq1)\ .$$
A: Here is a plot of Christian Blatter's bounds $\pm (n+1)/n^2$ and $a_n$:

I posted a new MSE question here about the convergence of series $\sum_{n=1}^{\infty}a_n$.
A: Alex's suggestion that the sequence is bounded by zero is incorrect.  The graph disproves this.  
Kelenner's suggestion is at best confusing.  He needs absolute values in multiple places.  His recursive case has not been proven.
m seems to serve two different purposes-
i.  a_n≥ m  but later 
ii. a_m> 0
Feel free to edit with LaTeX, etc.
A: Compute first $a_1$. I have found $a_1=-0,3689...$. Now use $|a_n|\leq \frac{n+1}{n^2}\leq \frac{2}{n}$. We have $\frac{2}{n}<0,3689...$ for $n\geq 6$. Hence, for $n\geq 6$, we have $a_n>a_1$. Now you see that if $m$ is the minimum of $a_k$, $k=1,2,3,4,5$, we have $a_n\geq m$ for all $n$, and there exists $k\leq 5$ such that $a_k=m$. Hence $m$ is the lower bound you want.
For the upper bound, compute $a_2,a_3,a_4,a_5,...$ and find a $m$ such that $a_m>0$, and use the same way of proof.
