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$$a_n=\frac{n\cos(n)+\sin(n^2+2n-5)}{n^2}$$

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.

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what about using $|cos(n)|\leq 1$ and $|\sin(\cdot)|\leq 1$? By "the bound of this sequence," do you mean the limit? –  Michael Sep 4 at 6:42
    
then i get $\frac{n+1}{n^2}$ all examples I have seen are basic. –  Parhs Sep 4 at 6:45
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I am confused: Do you not like $(n+1)/n^2$ for some reason? It is not really clear what you are looking for. –  Michael Sep 4 at 6:50
    
I dont know how to find the bound from that –  Parhs Sep 4 at 6:52
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So, you don't like $(n+1)/n^2$ because it is not a real number? Then you should find a real number that is greater than that, for all $n \in \{1, 2, 3, \ldots\}$. –  Michael Sep 4 at 6:56

4 Answers 4

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 $$|a_n|\leq{n+1\over n^2}\qquad(n\geq1)\ .$$

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Very nice results! –  mike Sep 4 at 8:48

Here is a plot of Christian Blatter's bounds $\pm (n+1)/n^2$ and $a_n$: enter image description here

I posted a new MSE question here about the convergence of series $\sum_{n=1}^{\infty}a_n$.

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so it is correct to say that $|a_n|<1$ (boundary) –  Parhs Sep 4 at 10:39

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.

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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.

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