Thread: Another summation (or geometric?) Series

Hello everyone,

I need the formula of the following series:

_{i = 0}∑^{k - 1} n/[(log n) - i] = ? for all n != 0 and n != 2ⁱ

Thanks!

Originally Posted by mohamedennahdi

Hello everyone,

I need the formula of the following series:

_{i = 0}∑^{k - 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?

for

Yes.

Wolfram gets this.

This one is tough!

Is there any one to agree that:

_i=0∑^{(log n - 1)} n / ((log2 n) - i) = n (_j=1∑^{log n} 1 / j) ?

Originally Posted by mohamedennahdi

Is there any one to agree that:

_i=0∑^{(log n - 1)} n / ((log2 n) - i) = n (_j=1∑^{log n} 1 / j) ?

No I completely disagree. The counter in the sum can only take integer values.

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