Struggling to show that this series converges.

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In summary: The left side of the inequality diverges (because 0<= a <= 1) and so the series on the right side diverges.
  • #1
Set Abominae
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Hi there,
I'm trying to show that [tex]\sum\frac{log(r)}{r^{a}}[/tex] (r going from 1 to infinity) converges using the ratio test, but I can't seem to deal with the log(r+1)/log(r) term.
Any help would be much appreciated.

Thanks.
 
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  • #2
Is it not sufficient to note that r^a/(r+1)^a is a null sequence, so anything it is multiplied by will also be a null sequence? (null x bounded = null...etc).
 
  • #3
i would use this (it is supposed that your series goes from n=1 to infinite

integral criteria , if the integral [tex] \int _{1}^{\infty} dx log(x)x^{-a} [/tex] converges so does your series

on the other hand an useful trick is to use the expansion of log (r+1) for big r

log(r+1)=log(r)+log(1+1/r) and since 1/r is small you can expand it into a/r+b/r*r and so on
 
  • #4
It depends on your value of a. If a is negative, the series diverges. If 0<= a <= 1 then for r >=3 we have [tex] \frac{1}{r^a} < \frac{ln(r)}{r^a} [/tex] and so [tex] \sum\frac{1}{r^{a}} < \sum\frac{ln(r)}{r^{a}} [/tex]. The left side of the inequality diverges (because 0<= a <= 1) and so the series on the right side diverges.

I haven't thought about a > 1 yet. I'll get back to that when I get back home.
 
  • #5
JG89 said:
It depends on your value of a. If a is negative, the series diverges. If 0<= a <= 1 then for r >=3 we have [tex] \frac{1}{r^a} < \frac{ln(r)}{r^a} [/tex] and so [tex] \sum\frac{1}{r^{a}} < \sum\frac{ln(r)}{r^{a}} [/tex]. The left side of the inequality diverges (because 0<= a <= 1) and so the series on the right side diverges.

I haven't thought about a > 1 yet. I'll get back to that when I get back home.

I should have mentioned, I was considering a>1. Thanks.
 

1. What is the definition of convergence in a series?

The definition of convergence in a series is the property of a series where the sum of its terms approaches a finite value as the number of terms increases. In other words, the series converges when the terms become smaller and smaller as more terms are added, ultimately leading to a final sum.

2. How can I determine if a series converges or diverges?

To determine if a series converges or diverges, you can use various convergence tests such as the comparison test, ratio test, root test, or integral test. These tests compare the given series to a known series with known convergence properties to determine its behavior.

3. Can a series converge to more than one value?

No, a series can only converge to one value. If a series has more than one possible limit, then it is considered to be divergent.

4. What is the difference between absolute and conditional convergence?

Absolute convergence occurs when a series converges regardless of the order of its terms, while conditional convergence only occurs when the terms are arranged in a specific order. In other words, a series with absolute convergence will always converge to the same value, while a series with conditional convergence may converge to different values depending on the order of its terms.

5. What is the significance of showing that a series converges?

The significance of showing that a series converges is that it allows us to make predictions and approximations in various mathematical and scientific applications. Convergent series are also easier to work with and manipulate, making them more useful in mathematical calculations and analyses.

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