- #1

Dr Zoidburg

- 39

- 0

Test for convergence/divergence:

[tex]\sum_{n=1}^{\infty} \frac {(n+1)}{n^3 ln(n+2)}[/tex]

What test should I do here? Can I rearrange the equation to be:

[tex] \frac{(\frac{1}{n^2} + \frac{1}{n^3})}{ln(n+2)}[/tex]

and then use L'Hopital's Rule?

Or do I need to use the Integral test? If so, what should I put u=?

Next Question:

For what values of k isthe following series absolutely convergent? For what values k>=0 conditionally convergent?

[tex]\sum_{n = 3}^{\infty}\frac{(-1)^n}{n(\ln{n})[\ln{(\ln{n})}]^k}[/tex]

I've spent a looooonnnng time on this! This is what I've got thus far:

make [tex]u= ln(ln(n))[/tex]

[tex]du=\frac{1}{n(ln(n))} dn[/tex]

And then integrate as follows:

[tex]\lim_{b = 3 \to \infty} \int_{3}^{b} \frac{1}{u^k} du[/tex]

which gives us:

[tex]\frac{u^{(1-k)}}{1-k}|_{3}^{b}[/tex]

So does this mean it's convergent for k>1 and divergent for k=1?

Last question:

prove:

[tex]\lim_{n \to \infty} \frac {n^p}{a^n} = 0[/tex] where a & p are constants and a>1 & p>0

I've tried using the Ratio test here, but I ended up with 1/a, which obviously is not 0.

Thanks for any and all help/advice!