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Numbers for which it coverges

  • Thread starter zeion
  • Start date
  • #1
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Homework Statement



Find the positive numbers r for which [tex]\sum \frac{r^k}{k^r}[/tex] converges.

Homework Equations





The Attempt at a Solution



What method do I use for this?

[tex]\frac {a_{k+1}}{a_k} = \frac {r^{k+1}}{(k+1)^r} \cdot \frac{k^r}{r^k}
= r \cdot (\frac{1}{k})^{\frac{r}{k}} [/tex]
 

Answers and Replies

  • #2
33,494
5,186

Homework Statement



Find the positive numbers r for which [tex]\sum \frac{r^k}{k^r}[/tex] converges.

Homework Equations





The Attempt at a Solution



What method do I use for this?

[tex]\frac {a_{k+1}}{a_k} = \frac {r^{k+1}}{(k+1)^r} \cdot \frac{k^r}{r^k}
= r \cdot (\frac{1}{k})^{\frac{r}{k}} [/tex]
The second and third expressions aren't equal...
 
  • #3
467
0
Ok so I have [tex] \frac {a_{k+1}}{a_k} = \frac {r^{k+1}}{(k+1)^r} \cdot \frac{k^r}{r^k}
= r \cdot \frac{k^r}{(k+1)^r } = r \cdot (\frac{k}{(k+1)})^r[/tex] ?
 
  • #4
467
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then [tex]r \cdot (1 - \frac{1}{k+1})^r \to r \cdot 1^r [/tex] ?
 
  • #5
467
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then r needs to be < 1?
 
  • #6
33,494
5,186
What if r = -200? That's less than 1.
 
  • #7
467
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I thought if the ratio is < 1 then it converges?
 
  • #8
33,494
5,186
The ratio test is usually given in terms of absolute values. IOW,
[tex]\lim_n \frac{|a_{n + 1}|}{|a_n|}[/tex]

For this problem it's given that r is a positive number, so since the ratio of those terms is less than 1, the series converges.
 
  • #9
467
0
So how do I know the ratio is less than 1?
 
  • #10
33,494
5,186
I thought you already knew that. What I thought you did was to evaluate this limit:
[tex] \lim_{k \to \infty} \frac {a_{k+1}}{a_k} = \lim_{k \to \infty} \frac {r^{k+1}}{(k+1)^r} \cdot \frac{k^r}{r^k}= \lim_{k \to \infty} r \cdot \frac{k^r}{(k+1)^r } = \lim_{k \to \infty} r \cdot (\frac{k}{(k+1)})^r[/tex]
What do you get for this limit? And what does that value imply for the test you are using?
 
  • #11
467
0
I get [tex]r \cdot 1^r = r[/tex]?
And r is positive
 
  • #12
33,494
5,186
zeion said:
Find the positive numbers r for which [tex]\sum \frac{r^k}{k^r}[/tex]
converges.
From your work, what are the values of r that cause this series to converge?
 
  • #13
467
0
I need the ratio to approach something < 1?
So since r is positive I need it to be 0 < r < 1?
 
  • #14
33,494
5,186
C'mon, show some confidence. Tell me, don't ask me.
 
  • #15
33,494
5,186
I need the ratio to approach something < 1?
So since r is positive I need it to be 0 < r < 1?
C'mon, show a little confidence in your ability. Tell me, don't ask me.
 

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