Infinite alternating series with repeated logarithms

LeifEricson
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Homework Statement



Calculate the sum of the following series:
\sum_{i=2}^{\infty}(-1)^i \cdot \lg ^{(i)} n

Where (i) as a super-script signifies number of times lg was operated i.e. \lg ^{(3)} n = (\lg (\lg (\lg n))), and n is a natural number.

Homework Equations





The Attempt at a Solution



I proved by Leibniz test that this series converges. But I don't know how to find the number to which it converges to.

Thanks.
 
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I really don't know how to help you... but since you know it converges you can use the numerical estimate to see if it looks like any famous number?

I hope someone helps I'm really interested in this!
 
If you try doing a numerical estimate it will blow up. Which is informative. Eventually ln^(k)(n)<1. Then ln^(k+1)(n)<0. Now what's ln^(k+2)(n)?? The series isn't even well defined much less convergent.
 
You are right.
The series isn't well defined so I cancel this question. I apologize.
 
Oh that makes me feel a little better. I could not even see how you had it converging.

:-p
 
LeifEricson said:
You are right.
The series isn't well defined so I cancel this question. I apologize.

No need for apologies! It's a perfectly valid question. It makes a good 'think about it' exercise. It does look at first glance like it ought to be a good candidate for an alternating series test.
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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