# Thread: divergent or convergent but turns out undefined?

1. ## divergent or convergent but turns out undefined?

I am trying to find divergence or convergence using Maple of;

SUM from n = 1 to inf of the equation:

(1+sin(n)/n)^2

I run it in Maple as a ratio test but when I try to take the limit, I get that the solution is undefined?

2. Originally Posted by orendacl
I am trying to find divergence or convergence using Maple of;

SUM from n = 1 to inf of the equation:

(1+sin(n)/n)^2

I run it in Maple as a ratio test but when I try to take the limit, I get that the solution is undefined?
You can't evaluate $\lim_{n \to \infty}\frac{\sin{n}}{n}$, so I don't believe that you can determine convergence or divergence using the ratio test...

3. Originally Posted by orendacl
I am trying to find divergence or convergence using Maple of;

SUM from n = 1 to inf of the equation:

(1+sin(n)/n)^2

I run it in Maple as a ratio test but when I try to take the limit, I get that the solution is undefined?
In fact, by the simple divergence test...

$\lim_{n \to \infty}\left(1 + \frac{\sin{n}}{n}\right)^2 = 1$. Since this does not $\to 0$, the series is divergent.

4. My apologies.
I really need to figure out how to get the fancy equations in these postings.
The actual question is:

[(1+sin(n))/n]^2

Not:

(1+sin(n)/n)^2

Now, that probably changes things!
I agree, that in the last scenario where "1" was not part of the numerator...then it would be divergent but what if we use the new equation - again, my apologies for lack of expertise in posting the question correctly the first time.

5. Originally Posted by Prove It
You can't evaluate $\lim_{n \to \infty}\frac{\sin{n}}{n}$,
Use the sandwich theorem!

6. Originally Posted by orendacl
I really need to figure out how to get the fancy equations in these postings.
Why not learn to post in symbols? You can use LaTeX tags.
$$\left[\frac{1+\sin(x)}{n}\right]^2$$ gives $\left[\frac{1+\sin(x)}{n}\right]^2$.

7. OK, here it goes:

$\left[\frac{1+\sin(n)}{n}\right]^2$

that is my equation:

I get this as being undefined or that...there is no convergence or divergence for this problem?

8. Note that $\left[1+\sin(n)\right]^2\le4$.
So $\sum\limits_{n = 1}^\infty {\left[ {\frac{{1 + \sin (n)}}
{n}} \right]^2 } \leqslant \sum\limits_{n = 1}^\infty {\frac{4}
{{n^2 }}}$

9. Originally Posted by Prove It
You can't evaluate $\lim_{n \to \infty}\frac{\sin{n}}{n}$, so I don't believe that you can determine convergence or divergence using the ratio test...
What? since sin(x) is always between -1 and 1 and n goes to infinity, $\lim_{n\to \infty}\frac{sin(n)}{n}= 0$

10. I agree...You "can" evaluate that limit, and it does = 0

11. I think this goes infinity

12. Originally Posted by HallsofIvy
What? since sin(x) is always between -1 and 1 and n goes to infinity, $\lim_{n\to \infty}\frac{sin(n)}{n}= 0$
Thanks, I assumed that since $\lim_{n \to \infty}\sin{n}$ doesn't exist, due to its oscillatory nature, then neither would $\lim_{n \to \infty}\frac{\sin{n}}{n}$.

What happens at the points where you end up with $\frac{0}{\infty}$ though?

13. Originally Posted by Prove It
Thanks, I assumed that since $\lim_{n \to \infty}\sin{n}$ doesn't exist, due to its oscillatory nature, then neither would $\lim_{n \to \infty}\frac{\sin{n}}{n}$.

What happens at the points where you end up with $\frac{0}{\infty}$ though?
That still 0- you have very very small numbers divided by very very large numbers- even closer to 0!

14. Yes but,
The terms of our series are strictly between 0 and 4/n^2.
"both of those series" WILL converge so shouldn't we then just use the comparison test?

15. Originally Posted by orendacl
Yes but,
The terms of our series are strictly between 0 and 4/n^2.
"both of those series" WILL converge so shouldn't we then just use the comparison test?
Absolutely correct.
The series converges by the basic comparison test.