Convergence of a series with tests

[trap101]

Determine whether the series converges or diverges:

$\sum$ ln k/ k³

now I said that this series converges by the comparison test, using ln k / k since I know that goes to 0

Would that be the right logic?

 

[DryRun]

In the series, k tends towards what value? Your question is missing that information.

 

[trap101]

oh, sorry. k is tending towards ∞

 

[LCKurtz]

Determine whether the series converges or diverges:

$\sum$ ln k/ k³

now I said that this series converges by the comparison test, using ln k / k since I know that goes to 0

Would that be the right logic?

You haven't written enough to tell. What comparison test are you using? What convergent series are you comparing it with?

 

[trap101]

You haven't written enough to tell. What comparison test are you using? What convergent series are you comparing it with?

I was using a basic comparison test. And I was using ln k/ k as my convergent series. Unless ln k/ k is not a convergent series, but it is a convergent sequence though.

 

[LCKurtz]

I was using a basic comparison test. And I was using ln k/ k as my convergent series. Unless ln k/ k is not a convergent series, but it is a convergent sequence though.

Maybe you confuse sequences and series. What makes you think$$
\sum_{k=1}^\infty \frac {\ln k}{k}$$is a convergent series? Just because the kth term $\frac {\ln k} k\rightarrow 0$ doesn't mean the series converges.

 

[trap101]

Maybe you confuse sequences and series. What makes you think$$
\sum_{k=1}^\infty \frac {\ln k}{k}$$is a convergent series? Just because the kth term $\frac {\ln k} k\rightarrow 0$ doesn't mean the series converges.

Actually I've been trying to straighten out that confusion, but your right I do mix them up sometimes. Ok, well by a comparison test with 1/k, that would mean $$
\sum_{k=1}^\infty \frac {\ln k}{k}$$ is a divergent series

but then that doesn't help me here because $$
\sum_{k=1}^\infty \frac {\ln k}{k}$$

is larger than
$$
\sum_{k=1}^\infty \frac {\ln k}{k^3}$$

perhaps a limit comparison with

$$
\sum_{k=1}^\infty \frac {\ln k}{k}$$ would work?

 

[micromass]

Use that ln(k)≤k for large k.

 

[trap101]

Use that ln(k)≤k for large k.

Yes, then that simplifies to 1/k² which converges based off of p-series.

Thanks

 

[trap101]

On another note...what's a good way to stop confusing sequences, partial sums, and series?

 

[micromass]

On another note...what's a good way to stop confusing sequences, partial sums, and series?

Concentration.

 

[DryRun]

Concentration.

 

[trap101]

I have another quick question about the basic convergence test:


If lim a_n as n-->∞ ≠ 0 then Ʃ a_n will diverge.



is the a_n that is being talked about at the beginning of the theorem the "sequence" a_n or is it the values of the partial sum of all a_n that they are talking about?

 

[the_kid]

I have another quick question about the basic convergence test:If lim a_n as n-->∞ ≠ 0 then Ʃ a_n will diverge.
is the a_n that is being talked about at the beginning of the theorem the "sequence" a_n or is it the values of the partial sum of all a_n that they are talking about?

The a$_{n}$ you are referring to is a sequence of terms. The k-th partial sum of $\sum$a$_{n}$ is defined S$_{k}$=a$_{0}$+a$_{1}$+a$_{2}$+...+a$_{k}$. I hope this clears up some of the confusion you've been having regarding series, sequences, and partial sums.

 

[LCKurtz]

I have another quick question about the basic convergence test:If lim a_n as n-->∞ ≠ 0 then Ʃ a_n will diverge.
is the a_n that is being talked about at the beginning of the theorem the "sequence" a_n or is it the values of the partial sum of all a_n that they are talking about?

$a_n$ is the nth term of the series $\sum_{n=1}^\infty a_n$. Of course if the series starts with $n=0$ it isn't exactly the nth term any more, it's the "general term", but it is still a good way to keep things straight.