How do I show that diverges? Also, how do I show that converges? If you're too lazy for a full response, I think that just naming a test would be enough for such a problem.
Any input would be greatly appreciated!
Thanks in advance!
How do I show that diverges? Also, how do I show that converges? If you're too lazy for a full response, I think that just naming a test would be enough for such a problem.
Any input would be greatly appreciated!
Thanks in advance!
Use the integral test for both
$\displaystyle \int\limits_2^\infty\frac{8\,dx}{3x\ln x}=\frac{8}{3}\lim\limits_{b\to\infty}\int\limits_ 2^b\frac{d(\ln x)}{\ln x}=\left.{\frac{8}{3}\lim\limits_{b\to\infty}\ln\l n x}\right|_2^b =$ $\displaystyle \frac{8}{3}\lim\limits_{b \to\infty}\left(\ln\ln b - \ln \ln 2\right) = \infty.$
$\displaystyle \int\limits_1^\infty {xe^{-x^2}\,dx}=-\frac{1}{2}\mathop {\lim }\limits_{b\to\infty} \int\limits_1^b{e^{-x^2}\,d\left(-x^2\right)}= \left. { - \frac{1}{2}\lim\limits_{b\to\infty}e^{-x^2}}\right|_1^b =$ $\displaystyle -\frac{1}{2}\lim\limits_{b\to\infty}\left(e^{-b^2}-e^{-1}\right) = \frac{e^{-1}}{2}.$
when you use a u-substution, you have:
$\displaystyle \int \frac{8}{3n.ln(n)} dx$
let $\displaystyle u = ln(n) \implies du = \frac{1}{n} dn \implies dn = n.du$
so,
$\displaystyle \frac{8}{3} \int \frac{n}{n. u} du = \frac{8}{3} \int{1}{u} du = \frac{8}{3} log(u) = \frac{8}{3} ln(ln(n))$
you can use the limits given to see that the integral diverges.. this has been done in the above post by DeMath
For substitution $\displaystyle u=\ln{x}$ :
If $\displaystyle x=2$ then $\displaystyle u=\ln{2}$
If $\displaystyle x\to\infty$ then also $\displaystyle u\to\infty$
I.e. you have
$\displaystyle \frac{8}{3}\int\limits_2^\infty\frac{dx}{x\ln{x}}= \left\{\begin{gathered}u=\ln{x}\hfill\\du=\frac{dx }{x}\hfill\\\end{gathered}\right\}=\frac{8}{3}\int \limits_{\ln2}^\infty\frac{du}{u}$