Hello everyone, I need the formula of the following series: _{i = 0}∑^{k - 1 }n/[(log n) - i] = ? for all n != 0 and n != 2^{i} Thanks!
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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 != 2^{i} 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.
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) ?
Last edited by mohamedennahdi; Jul 26th 2012 at 07:33 PM.
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).
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