1. ## Series problems

I need some help with these series problems, i do not know what test to use on them. Help appreciated thanks (also i do not know how to type a summation symbol).

Summation n=0 to infinity 13arctan(n)

summation n=1 to infinity sin(43n) / 1+n^2

2. #1: Use the divergence test. If $\lim_{n \to \infty} a_n \neq 0$, then the $\sum a_n$ diverges.

#2: Use the fact that if $\sum |a_n|$ converges, then so does $\sum a_n$. Then make use of the comparison test by using $|\sin x | \leq 1$

3. $

\lim_{n \to \infty} 13arctan(n)
$
does not go to 0 so divergent(wow simple).

I know this probebly isent right but using the direct comparison test the best i could come up with is
$
\sin (43n)/1+n^2 \le 2/n^2
$

less then a p series that is convergent meaning convergent?

4. #1:

#2: You can't use a direct comparison because one of the conditions of the comparison test is that the terms of your sequence has to be positive for all integers after some $n$th term.

This is why we consider the sequence $|a_n|$ where all terms are obviously positive and is why the theorem I gave you is useful.

So: $\sum \left| \frac{\sin (43n)}{1+n^2} \right| = \sum \frac{|\sin (43n)|}{1+n^2} {\color{red} \ \leq \ } \sum \frac{1}{1+n^2} < \cdots$

Can you finish?

5. I would then show it is less then a convergent p series yes, but is that really less or equal because it has a numerator which would make it larger then the one without one... wouldn't it?

6. Think about it this way: $|\sin (43n)| \leq 1$

Multiply both sides by $\frac{1}{1+n^2} > 0$ to get: $\frac{|\sin (43n)|}{1+n^2} \leq \frac{1}{1+n^2}$

7. Thanks, i realize sin(x) will never be greater then one so that would be possible.