Is there a way to rewrite these following expressions so that they have $\displaystyle (-1)^n\cdot$ in front of them?
$\displaystyle \frac{\sin(n)+\cos(n)}{n^2}$
$\displaystyle (-2)^n\frac{1}{n^2+7}$
Is there a way to rewrite these following expressions so that they have $\displaystyle (-1)^n\cdot$ in front of them?
$\displaystyle \frac{\sin(n)+\cos(n)}{n^2}$
$\displaystyle (-2)^n\frac{1}{n^2+7}$
Are you sure the first one can't be rewritten?
I need to show that $\displaystyle \sum^\infty_{n=1}\frac{\sin(n)+\cos(n)}{n^2}$ is either convergent or divergent and since the expression is an alternating series, I should be able to use Leibniz criterion however I need to rewrite it with $\displaystyle (-1)^n$ as a factor.
Are you sure this can't be done?
Since $\displaystyle \frac{\sqrt{2}}{n^2}$ is convergent, $\displaystyle \frac{\sin(n)+\cos(n)}{n^2}$ must also be convergent.
But how to test if it is absolute convergent? Isn't there any ways to use the alternating series test? And how did you come up with the $\displaystyle \sqrt{2}$?
Well, there's your first problem. This isn't an alternating series. $\displaystyle \frac{sin(1)+ cos(1)}{1^2}= 1.38177$ and $\displaystyle \frac{sin(2)+ cos(2)}{2^2}= 0.12328$ both of which are positive
, I should be able to use Leibniz criterion however I need to rewrite it with $\displaystyle (-1)^n$ as a factor.
Are you sure this can't be done?
I've always called it the alternating series test instead of Leibniz criterion - just in case you run across that terminology.
For absolute convergence, you can just say that:
$\displaystyle \left|\frac{\sin(n)+\cos(n)}{n^2}\right| \le \frac{|\sin(n)|+|\cos(n)|}{n^2} \le \frac{2}{n^2}$
It's true that skeeter's bound is tighter, but you don't need it to prove convergence.
- Hollywood