2

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

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.

JimmyK4542
  • 54,331
GorillaApe
  • 1,071
  • 1
    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 04 '14 at 06:42
  • then i get $\frac{n+1}{n^2}$ all examples I have seen are basic. – GorillaApe Sep 04 '14 at 06:45
  • 1
    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 04 '14 at 06:50
  • I dont know how to find the bound from that – GorillaApe Sep 04 '14 at 06:52
  • Wow. Are you intentionally trying to be unclear? What do you mean by "the bound" ?? – Michael Sep 04 '14 at 06:53
  • I have to prove that sequence is bounded below and above. But dont know how to do it. – GorillaApe Sep 04 '14 at 06:54
  • 1
    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 04 '14 at 06:56
  • Yes I thought that it should be a real number. However I found 2 as $(n+1)/n^2$ seems decreasing. – GorillaApe Sep 04 '14 at 07:02
  • I posted a new MSE question about the convergence of this series $\sum_{n=1}^{\infty}a_n$.

    http://math.stackexchange.com/questions/919326/convergence-of-series-sum-n-1-infty-a-n-with-a-n-le-fracn1n2

     – mike Sep 04 '14 at 13:47

4 Answers4

3

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

Christian Blatter
  • 226,825
2

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 $\sum_{n=1}^{\infty}a_n$.

Community
  • 1
mike
  • 5,604
1

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.

nickalh
  • 1,263
0

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.

Kelenner
  • 18,734
  • 26
  • 36