Not sure how to proceed with this series

[frasifrasi]

With this series,

Summation from 2 to infinity of

(-1)^(n).n/ln(n)

--> Am I supposed to do the alternating series test?

If I do, lim n --> infinity n/ln(n): am I supposed to do l'hopital here or is there a more straight foward method?


Thank you.

 

[Office_Shredder]

L'hopital is a pretty good way to do \frac {n}{ln(n)} as n goes to infinity

 

[frasifrasi]

ok, it goes to infinity, but the first part of the alternating series, how can I determined that \frac {n+1}{ln(n+1)}?

also, for a different series altogether, I am not sure how \frac {(-1 to (n-1))}{2n} --> -1^(n-1)
is conditionally convergent. If I take the abs value, I get 1/2n, which is a p series and diverges. And the alternating series diverges, so would it not diverge?

 

[Office_Shredder]

To be completely honest, I have no idea what you're trying to ask. Did you mess up the latex?

The fact that the limit goes to infinity specifically means it doesn't go to zero, which is a necessary condition for convergence

 

[frasifrasi]

but the first series I posted is supposed to be conditionally convergent. Can anyone else help?

 

[Avodyne]

but the first series I posted is supposed to be conditionally convergent.

It isn't.

 

[frasifrasi]

That is what it says in the answer key, condinionally...

Can anyone else comment?

 

[Dick]

That is what it says in the answer key, condinionally...

Can anyone else comment?

Divergent. The key is simply wrong or the problem is misstated.

 

[HallsofIvy]

Are you talking about the same series?

The "last series" referred to, I think, was
\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}
an "alternating series". That converges if and only if the associated sequence converges to 0 which is true. Of course, the series of absolute values is
\sum_{n=1}^\infty \frac{1}{n}
which does not converge so the sum converges "conditionally", not "absolutely".


For the first series
\sum_{n= 1}^\infty \frac{n}{ln n}[/itex]<br /> Since the associated sequence n/ln n does not go to 0 it cannot converge either conditionally nor absolutely.

 

[frasifrasi]

so, \sum_{n= 1}^\infty \frac{n}{ln n} --> the lim n-->infinity =

1/ 1/x = infinity, so it diverges, correct?

 

[Office_Shredder]

so, \sum_{n= 1}^\infty \frac{n}{ln n} --> the lim n-->infinity =

1/ 1/x = infinity, so it diverges, correct?

This would be better stated as
1/(1/n) -> infinity as n-> infinity, so n/ln(n) -> infinity as n->infinity by L'Hopital. Hence, the sum is divergent, as the summation terms do not tend to zero

 

[frasifrasi]

What about \sum_{n= 1}^\infty \frac{6^n + 5^n}{30^n}


- it's been a while, what method should I use for this?

 

[frasifrasi]

anyone (I am trying to prepare for a test : ) !)

 

[Avodyne]

Can you do \sum_{n= 1}^\infty x^n for |x|&lt;1?

 

[VietDao29]

What about \sum_{n= 1}^\infty \frac{6^n + 5^n}{30^n}


- it's been a while, what method should I use for this?

Big hint of the day: You can split it into 2 small parts, and solve it easily using the sum of geometric series.

Can you go from here? :)