Thursday, 30 June 2016

special functions - Bounds on geometric sum



Consider the sum x=1logxzx. We can assume that z1 (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 xn+1 we have ln(n+1)ln(x)ln(n+1)+xn1n+1 so
nx=1ln(x)zx+zln(n+1)zn+2zn+1x=1ln(x)zxnx=1ln(x)zx+1+(z1)(n+1)ln(n+1)(n+1)(z1)2zn



Some explanation for the PolyLog: by definition
PolyLog(p,1/z)=x=1xpzx


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
pxp=xplog(x)



No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...