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Series Convergence and Divergence test!

  • #1

Homework Statement



So my question was Sum- (n=2) ln(n)/n

Homework Equations



I noticed that you can only limit comparison, because so far, I have tried doing all the other test such as the nth term test, p-series, integral(i have no idea how to integrate that).

The Attempt at a Solution

 

Answers and Replies

  • #2
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I guess you mean
$$\sum_{n=2}^\infty \frac{\ln(n)}{n}$$
Limit comparison is good. Do you expect it to converge or not? What did you try so far as comparisons?
 
  • #3
I guess you mean
$$\sum_{n=2}^\infty \frac{\ln(n)}{n}$$
Limit comparison is good. Do you expect it to converge or not? What did you try so far as comparisons?
Well, it is supposed to diverge? I don't even know what I'm supposed to be comparing it to
 
  • #4
34,046
9,896
Well, it is supposed to diverge? I don't even know what I'm supposed to be comparing it to
I'm sure you had convergent or divergent series which had something in common with this expression before. The most useful one is probably the textbook example of a series that gets analzed for convergence.
 
  • #5
I'm sure you had convergent or divergent series which had something in common with this expression before. The most useful one is probably the textbook example of a series that gets analzed for convergence.
I tried looking one up, they gave me the answer instead of an explanation or work, and that is the MOST important part.
 
  • #6
34,046
9,896
Then which part of the answer was unclear?
 
  • #7
Then which part of the answer was unclear?
Okay, so you know how to do a limit comparison test we need to compare the function to something right? I'm not sure what a comparison function would be for this function.
 
  • #8
Ray Vickson
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Okay, so you know how to do a limit comparison test we need to compare the function to something right? I'm not sure what a comparison function would be for this function.
We cannot give you more hints without essentially doing the question for you.
 
  • #9
We cannot give you more hints without essentially doing the question for you.
Let me try...so from what I recall, my teacher said that when we are finding a function to compare, we have to look the highest power from each the numerator and denominator of the function. If that is the case, wouldn't you compare it to ln(n)?
 
  • #10
34,046
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It is possible to do that comparison, but it won't tell you anything new.

Can you list series where you checked convergence before?
 
  • #11
So you want me to list out how I tested for convergence?
 
  • #12
34,046
9,896
Now how, but what. At the point where you get those homework problems, you should have tested a few series for convergence before. Can you list them (and the test result)?
 
  • #13
oh! Okay...
∑1/(ln2)^n
r = ιrι = ι1/ln 2 ι ≥1 so, divergent by geometric series.

P.S. how do you put the math thingies in this thing....I'm very new, and I'm confused. :/
 
  • #14
34,046
9,896
oh! Okay...
∑1/(ln2)^n
r = ιrι = ι1/ln 2 ι ≥1 so, divergent by geometric series.
That is certainly not the only one you had before.

P.S. how do you put the math thingies in this thing....I'm very new, and I'm confused. :/
With LaTeX.
 
  • #15
LCKurtz
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Another test you might consider is the integral test.
 
  • #16
Another test you might consider is the integral test.
Yes, but the integral test is barely used...
 
  • #17
That is certainly not the only one you had before.

With LaTeX.
That was literally the only one that I found...
 
  • #18
LCKurtz
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Another test you might consider is the integral test.
Yes, but the integral test is barely used...
I have no idea what do you mean by that. It easily works your problem. Have you tried it?

And along the lines of comparison tests, I don't think the "limit comparison test" is what you want anyway. If you think your series diverges try looking for a known divergent series that is smaller.
 

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