- #1
izen
- 51
- 0
Homework Statement
Using the limit comparison test to determine convergence or divergence
(1) [itex]\sum^{\infty}_{n=1} \frac{(ln(n))^{3}}{n^{3}}[/itex]
(2) [itex]\sum^{\infty}_{n=1} \frac{1}{\sqrt{n} ln(x)}[/itex]
(3)[itex]\sum^{\infty}_{n=1} \frac{(ln(n))^{2}}{n^{\frac{3}{2}}}[/itex]
Homework Equations
The Attempt at a Solution
I have problem find thing the another series (Bn) compare with An. First I choose Bn when 'n' grows large to represent An
The first question I chose Bn = [itex]\frac{1}{n^{3}}[/itex]
it is the p-series and p>1 so it converges
and then
[itex]lim_{n\rightarrow\infty} \frac{\frac{ln(n)^{3}}{n^{3}}}{\frac{1}{n^{3}}}[/itex]
[itex]lim_{n\rightarrow\infty} ln(n)^{3} = \infty[/itex]
which is wrong so I change to choose Bn = [itex]\frac{1}{n^{2}}[/itex]
[itex]lim_{n\rightarrow\infty} \frac{\frac{ln(n)^{3}}{n^{3}}}{\frac{1}{n^{2}}}[/itex]
[itex]lim_{n\rightarrow\infty} \frac{ln(n)^{3}}{n}= 0[/itex]
So I can conclude that An is convergent by limit comparison
but why I cannot use Bn = [itex]\frac{1}{n^{3}}[/itex] ?
=========================================
(2) and (3) are the same as my first question
(2) [itex]\sum^{\infty}_{n=1} \frac{1}{\sqrt{n} ln(x)}[/itex]
I chose Bn = [itex]\frac{1}{n^{1/2}}[/itex] but after I took the limit of An/Bn. I got 0. In the answer shows that I have to choose Bn = [itex]\frac{1}{n^{}}[/itex]
(3)[itex]\sum^{\infty}_{n=1} \frac{(ln(n))^{2}}{n^{\frac{3}{2}}}[/itex]
I chose Bn = [itex]\frac{1}{n^{3/2}}[/itex] after I took the limit of An/Bn. I got [itex]\infty[/itex]. In the answer shows that I have to choose Bn = [itex]\frac{1}{n^{5/4}}[/itex]
Please give some comments and advices
Thank you
Last edited by a moderator: