Consider the sum ∑∞x=1logxzx. We can assume that z≥1 (and is real). Mathematica gives this sum as
-PolyLog^(1, 0)[0,1/z]
Even after reading the manual page for PolyLog I don't understand what this function is like and I certainly don't know how the sum was derived.
Are there simple upper and lower bounds for this sum?
I also tried to compute ∫∞x=1logxzx in the hope that this would shed more light but Mathematica gives
Gamma[0, Log[z]]/Log[z]
which I also didn't find helpful.
Answer
I assume you mean ∞∑x=1ln(x)zx.
Since z>1 and ln(x) grows very slowly, the terms go to 0 rapidly, so you
can get good bounds by taking a partial sum and bounds for the "tail".
For x≥n+1 we have ln(n+1)≤ln(x)≤ln(n+1)+x−n−1n+1 so
n∑x=1ln(x)zx+zln(n+1)zn+2−zn+1≤∞∑x=1ln(x)zx≤n∑x=1ln(x)zx+1+(z−1)(n+1)ln(n+1)(n+1)(z−1)2zn
Some explanation for the PolyLog: by definition
PolyLog(p,1/z)=∞∑x=1x−pzx
Take the derivative of this with respect to p (which is what the (1,0) superscript refers to), evaluated at p=0 and you get −1 times your sum, because
∂∂px−p=−x−plog(x)
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