For the following summations from 1 to infinity, which tests for convergence would I use? I am confused on what to select.
1) 9/((n^2)*(sqrt(n)))
2) (-3*n*sqrt(n))/(n^2)
3) 3*(-1)^n
4) cos((n^5)+1)
5) (n^7)-10*n
Thanks
1) $\displaystyle \sum_{n=1}^{\infty}\frac{9}{n^2\sqrt{n}}\implies \sum_{n=1}^{\infty}\frac{9}{n^{\frac{5}{2}}}$. You can apply the integral test, or recognize that it's a p-series, where $\displaystyle \frac{5}{2}>1$ (thus, the series converges).
2) $\displaystyle \sum_{n=1}^{\infty}\frac{-3n\sqrt{n}}{n^2} \implies \sum_{n=1}^{\infty}-\frac{3}{n^{\frac{1}{2}}}$. Again apply the integral test, or recognize it's a p-series, where $\displaystyle \frac{1}{2}<1$ (thus, the series diverges).
3) I forget how to prove it, but it diverges. (expanding we get 3-3+3-3+3-3+...)
4) $\displaystyle \sum_{n=1}^{\infty}\cos(n^5+1)$. I believe you would use the limit comparison test here (?) : Thus,$\displaystyle \lim_{n\to{\infty}}cos(n^5) \ \text{DNE}$. Since this doesn't exist, then the series diverges. (I think this is correct)
5)$\displaystyle \sum_{n=1}^{\infty}n^7-10n$. I'm sure the integral test would work here as well. It's pretty evident that the series diverges.
Can someone verify these??
oh so if i put any number as i get bigger it has to go close to zero?
also for the integral test, after you integrate, what value to you use as upper bound?
like for example this says n^2 + 7:
Visual Calculus -
i dont quite get the step after they say the integral becomes....
EDIT: ahh alright i get it
$\displaystyle \frac{3^n}{8^n}=\left(\frac 38\right)^n$
This is the general term for a geometric sequence.
$\displaystyle \sum_{n=0}^\infty \left(\frac 38\right)^n=\lim_{n \to \infty} \frac{1-\left(\frac 38\right)^n}{1-\frac 38}$
It converges because $\displaystyle \frac 38 <1 \implies \lim_{n \to \infty} \left(\frac 38\right)^n=0$