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
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
#1: Use the divergence test. If $\displaystyle \lim_{n \to \infty} a_n \neq 0$, then the $\displaystyle \sum a_n$ diverges.
#2: Use the fact that if $\displaystyle \sum |a_n|$ converges, then so does $\displaystyle \sum a_n$. Then make use of the comparison test by using $\displaystyle |\sin x | \leq 1 $
$\displaystyle
\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
$\displaystyle
\sin (43n)/1+n^2 \le 2/n^2
$
less then a p series that is convergent meaning convergent?
#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 $\displaystyle n$th term.
This is why we consider the sequence $\displaystyle |a_n|$ where all terms are obviously positive and is why the theorem I gave you is useful.
So: $\displaystyle \sum \left| \frac{\sin (43n)}{1+n^2} \right| = \sum \frac{|\sin (43n)|}{1+n^2} {\color{red} \ \leq \ } \sum \frac{1}{1+n^2} < \cdots$
Can you finish?