Can someone help me with these series problems?

**s3a** · May 1st 2010, 07:46 AM

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!

**Laurent** · May 1st 2010, 08:31 AM

Originally Posted by s3a

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!

For the first one, you can use comparison with an integral (note that $\int\frac{dx}{x\ln x}=\ln(\ln x)$ ). (or Cauchy condensation test)

For the second one, convergence is very quick, so any basic test will work.

**DeMath** · May 1st 2010, 08:32 AM

Originally Posted by s3a

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

$\int\limits_2^\infty\frac{8\,dx}{3x\ln x}=\frac{8}{3}\lim\limits_{b\to\infty}\int\limits_ 2^b\frac{d(\ln x)}{\ln x}=\left.{\frac{8}{3}\lim\limits_{b\to\infty}\ln\l n x}\right|_2^b =$ $\frac{8}{3}\lim\limits_{b \to\infty}\left(\ln\ln b - \ln \ln 2\right) = \infty.$

$\int\limits_1^\infty {xe^{-x^2}\,dx}=-\frac{1}{2}\mathop {\lim }\limits_{b\to\infty} \int\limits_1^b{e^{-x^2}\,d\left(-x^2\right)}= \left. { - \frac{1}{2}\lim\limits_{b\to\infty}e^{-x^2}}\right|_1^b =$ $-\frac{1}{2}\lim\limits_{b\to\infty}\left(e^{-b^2}-e^{-1}\right) = \frac{e^{-1}}{2}.$

**s3a** · May 1st 2010, 11:04 AM

When I do the first one, I did it with a u substitution u = ln(n)
and my last step is 8/3 * lim t->inf ln(t) - ln(0) but I said ln(0) not ln(0.00000000000000000001...) which does not exist. So, what does that mean or did I do something wrong?

**harish21** · May 1st 2010, 11:44 AM

Originally Posted by s3a

When I do the first one, I did it with a u substitution u = ln(n)
and my last step is 8/3 * lim t->inf ln(t) - ln(0) but I said ln(0) not ln(0.00000000000000000001...) which does not exist. So, what does that mean or did I do something wrong?

when you use a u-substution, you have:

$\int \frac{8}{3n.ln(n)} dx$

let $u = ln(n) \implies du = \frac{1}{n} dn \implies dn = n.du$

so,

$\frac{8}{3} \int \frac{n}{n. u} du = \frac{8}{3} \int{1}{u} du = \frac{8}{3} log(u) = \frac{8}{3} ln(ln(n))$

you can use the limits given to see that the integral diverges.. this has been done in the above post by DeMath

**DeMath** · May 1st 2010, 11:47 AM

Originally Posted by s3a

When I do the first one, I did it with a u substitution u = ln(n)
and my last step is 8/3 * lim t->inf ln(t) - ln(0) but I said ln(0) not ln(0.00000000000000000001...) which does not exist. So, what does that mean or did I do something wrong?

For substitution $u=\ln{x}$ :

If $x=2$ then $u=\ln{2}$

If $x\to\infty$ then also $u\to\infty$

I.e. you have

$\frac{8}{3}\int\limits_2^\infty\frac{dx}{x\ln{x}}= \left\{\begin{gathered}u=\ln{x}\hfill\\du=\frac{dx }{x}\hfill\\\end{gathered}\right\}=\frac{8}{3}\int \limits_{\ln2}^\infty\frac{du}{u}$

**s3a** · May 1st 2010, 11:49 AM

Oh, my mistake was just that I thought the interval was from [1,inf] instead of [2,inf].

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