- #1
Dr Zoidburg
- 39
- 0
I've got a couple of problems I'm stuck on. Any help gratefully received!
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!
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!