1. ## Series C/D ?

hi

am stuck here =S

sigma(2-->infinity) 1 / (ln n)^9

2. Please any help for this ??
I belive it can be solved

3. Originally Posted by TWiX
hi

am stuck here =S

sigma(2-->infinity) 1 / (ln n)^9

Tonio

4. Originally Posted by tonio

Tonio
Thanks, but is it impossible to solve it using the standard tests ?
i saw it in an incomplete exam in my college =S
and cauchy's test is not covered in our material..

TWiX

5. Originally Posted by TWiX
Thanks, but is it impossible to solve it using the standard tests ?
i saw it in an incomplete exam in my college =S
and cauchy's test is not covered in our material..

TWiX

Not as far as I can tell right now, and the condensation test is pretty standard.

Tonio

6. Originally Posted by TWiX
is it impossible to solve it using the standard tests ?
Use basic comparsion.
$\displaystyle \begin{gathered} \ln \left( {\sqrt[9]{n}} \right) \leqslant \sqrt[9]{n} \hfill \\ \ln (n)^9 \leqslant 9^9 n \hfill \\ \frac{1} {{\ln (n)^9 }} \geqslant \frac{{9^{ - 9} }} {n} \hfill \\ \end{gathered}$

7. Originally Posted by Plato
Use basic comparsion.
$\displaystyle \begin{gathered} \ln \left( {\sqrt[9]{n}} \right) \leqslant \sqrt[9]{n} \hfill \\ \ln (n)^9 \leqslant 9^9 n \hfill \\ \frac{1} {{\ln (n)^9 }} \geqslant \frac{{9^{ - 9} }} {n} \hfill \\ \end{gathered}$
The power is on the "ln n" not just "n"

8. Originally Posted by TWiX
The power is on the "ln n" not just "n"
That is exactly what I wrote: $\displaystyle [ln(n)]^9=ln(n)^9=ln^9(n)$
That is perfectly standard notation.

If it were just on the n it would be $\displaystyle ln(n^9)$ which not what I wrote.

BTW: If you use a computer algebra system you must use $\displaystyle ln(n)^9$

9. Originally Posted by Plato
That is exactly what I wrote: $\displaystyle [ln(n)]^9=ln(n)^9=ln^9(n)$
That is perfectly standard notation.

If it were just on the n it would be $\displaystyle ln(n^9)$ which not what I wrote.
idk this :S
Thank you anyway ^_^

10. There is another solution
Limit comparison test with $\displaystyle b_n=\frac{1}{n^{\frac{9}{11}}}$
The limit $\displaystyle =\infty$
and $\displaystyle \sum b_n$ diverges.
Hence, Our series diverges.
It works, Right?

11. Originally Posted by TWiX
hi

am stuck here =S

sigma(2-->infinity) 1 / (ln n)^9
Originally Posted by tonio

Tonio
Originally Posted by Plato
Use basic comparsion.
$\displaystyle \begin{gathered} \ln \left( {\sqrt[9]{n}} \right) \leqslant \sqrt[9]{n} \hfill \\ \ln (n)^9 \leqslant 9^9 n \hfill \\ \frac{1} {{\ln (n)^9 }} \geqslant \frac{{9^{ - 9} }} {n} \hfill \\ \end{gathered}$
Originally Posted by TWiX
The power is on the "ln n" not just "n"
http://www.mathhelpforum.com/math-he...divergent.html
Did you think you'd get a different answer if you asked twice?

12. Originally Posted by Drexel28
http://www.mathhelpforum.com/math-he...divergent.html
Did you think you'd get a different answer if you asked twice?
Who told you am searching for different answer here ?
and where is the problem If I search for different answers ?
And in my last post here, I wrote another solution and I asked to check whether its correct or wrong, and you ignore that.
But where is the problem If I'm searching for different answers ?

13. Originally Posted by TWiX
Who told you am searching for different answer here ?
and where is the problem If I search for different answers ?
And in my last post here, I wrote another solution and I asked to check whether its correct or wrong, and you ignore that.
But where is the problem If I'm searching for different answers ?
Haha, I hope my last post didn't sound annoyed. I forgot to put "". What was your question?

14. Here is it

Originally Posted by TWiX
There is another solution
Limit comparison test with $\displaystyle b_n=\frac{1}{n^{\frac{9}{11}}}$
The limit $\displaystyle =\infty$
and $\displaystyle \sum b_n$ diverges.
Hence, Our series diverges.
It works, Right?

15. Originally Posted by TWiX
Here is it
Write explicitly what limit test you used.

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