# Thread: Another summation (or geometric?) Series

1. ## Another summation (or geometric?) Series

Hello everyone,

I need the formula of the following series:

i = 0k - 1 n/[(log n) - i] = ? for all n != 0 and n != 2i

Thanks!

2. ## Re: Another summation (or geometric?) Series

Originally Posted by mohamedennahdi
Hello everyone,

I need the formula of the following series:

i = 0k - 1 n/[(log n) - i] = ? for all n != 0 and n != 2i

Thanks!
Just so we're clear, is this the sum you're trying to evaluate?

\displaystyle \displaystyle \begin{align*} \sum_{i = 0}^{k - 1}{\frac{n}{\log{(n)} - i}} \end{align*} for \displaystyle \displaystyle \begin{align*} n \neq \left\{ 0, 2^i \right\} \end{align*}

Yes.

4. ## Re: Another summation (or geometric?) Series

Wolfram gets this.

5. ## Re: Another summation (or geometric?) Series

This one is tough!

6. ## Re: Another summation (or geometric?) Series

Is there any one to agree that:

i=0(log n - 1) n / ((log2 n) - i) = n (j=1log n 1 / j) ?

7. ## Re: Another summation (or geometric?) Series

Originally Posted by mohamedennahdi
Is there any one to agree that:

i=0(log n - 1) n / ((log2 n) - i) = n (j=1log n 1 / j) ?
No I completely disagree. The counter in the sum can only take integer values.

8. ## Re: Another summation (or geometric?) Series

You are right, the last suggestion is correct only when N is a power of the base (base 2 in my case).