Recall the polylogarithm $Li_s(z)=\sum_{n=1}^\infty \frac{z^n}{n^s}.$

For $|z|<1,$ is $\Re \left( \frac{Li_1(z)}{Li_2(z)} - \frac{2}{3}\frac{Li_2(z)}{Li_3(z)} \right)>0?$

The numerics suggest that it is true, but for the life of me I cannot figure out how to prove this.

  • $\begingroup$ Daniel, I am not convinced that your constant $2/3$ is sharp. It looks like (although I suggest you experiment yourself) if you replace it by $x$ ranging in $[0,1]$, then your real part attains it minimum at $z=-1$. $\endgroup$ Feb 7 '11 at 12:42
  • $\begingroup$ @Wadim: I am not convinced either. I did the numerics on the unit circle and got the same. The value attains its minimum at $z=-1.$ $\endgroup$ Feb 7 '11 at 13:27
  • 2
    $\begingroup$ @Daniel, so you could put a better constant instead: $108\zeta(3)\log(2)/\pi^4=0.92379318\dots$ with the equality attained at $z=-1$. I won't expect a nice proof... $\endgroup$ Feb 7 '11 at 13:46

The inequality seems to be true of the partial sums as well (though I haven't checked that thoroughly), so you might be able to prove it by induction, but I don't quite see how to do that.

Here's an idea for a different proof:

Convert to a common denominator (dropping a factor of 3):


Substitute the definition:

\[\def\sumty{\sum_{i=1}^\infty}\Re\left[\left(3\sumty\frac{z^n}{n}\sumty\frac{z^n}{n^3}-2\sumty\frac{z^n}{n^2}\sumty\frac{z^n}{n^2}\right)/\left( \sumty\frac{z^n}{n^2}\sumty\frac{z^n}{n^3}\right)\right]\]

Gather powers of $z$:

\[\def\sumk{\sum_{k=2}^\infty}\def\suml{\sum_{l=1}^{k-1}}\Re\left[\left( 3\sumk\left(\suml\frac{1}{(k-l)l^3}\right)z^k- 2\sumk\left(\suml\frac{1}{(k-l)^2l^2}\right)z^k \right)/\left( \sumk\left(\suml\frac{1}{(k-l)^2l^3}\right)z^k \right)\right]\]

Combine and convert to a common demoninator in the numerator:

\[\def\sumk{\sum_{k=2}^\infty}\def\suml{\sum_{l=1}^{k-1}}\Re\left[\left( \sumk\left(\suml\frac{3(k-l)-2l}{(k-l)^2l^3}\right)z^k \right)/\left( \sumk\left(\suml\frac{1}{(k-l)^2l^3}\right)z^k \right)\right]\]

Divide out the leading term $z^2$, and take a $1$ out of the sum (thereby removing the constant term in the numerator):

\[\def\sumkk#1{\sum_{k=#1}^\infty}\def\suml{\sum_{l=1}^{k-1}} 1 + \Re\left[\left( \sumkk3\left(\suml\frac{3(k-l)-2l-1}{(k-l)^2l^3}\right)z^{k-2} \right)/\left( \sumkk2\left(\suml\frac{1}{(k-l)^2l^3}\right)z^{k-2} \right)\right]\]

The coefficients in the numerator decay with $1/k$, the ones in the denominator with $1/k^2$. The leading terms are (courtesy of WolframAlpha):

\[ 1 + \Re\left[\frac{\frac{1}{2}z+\frac{475}{864}z^2+\frac{445}{864}z^3+...}{1+\frac{3}{8}z+\frac{155}{864}z^2+\frac{175}{1728}z^3+...}\right]\]

Note that the leading coefficients in the numerator are all close to $1/2$. Thus, if we multiply the numerator by $1-z$, the leading terms will nearly cancel, and the coefficients will decay with $1/k^2$, as in the denominator. Doing that (and also pulling out a factor of $z/2$ from the numerator) yields:

\[\def\sumkk#1{\sum_{k=#1}^\infty}\def\suml{\sum_{l=1}^{k-1}} 1 + \Re\left[\frac{z}{2(1-z)}\left( \sumkk3\left(2\Delta_k\suml\frac{3(k-l)-2l-1}{(k-l)^2l^3}\right)z^{k-3} \right)/\left( \sumkk2\left(\suml\frac{1}{(k-l)^2l^3}\right)z^{k-2} \right)\right]\]

where $\Delta_k f(k) := f(k) - f (k-1)$. The leading terms are now

\[ 1 + \Re\left[\frac{z}{2(1-z)} \frac{1+\frac{43}{432}z- \frac{5}{72}z^2+...}{1+\frac{3}{8}z+\frac{155}{864}z^2+\frac{175}{1728}z^3+...}\right]\]

I think from there you may be able to show by distinguishing cases and making use of the $1/k^2$ decay of the coefficients that the real part in the second term can never be less than or equal to $-1$. If the term $3/8z$ in the denominator is too big and causes trouble, it might help to multiply through by $1-3/8z$, as in the numerator.

I'd appreciate if you let me know what you're using this for :-)

  • 1
    $\begingroup$ It has some connection to the theory of plane partitions. Its a bit off topic to discuss here but if your interested message me. $\endgroup$ Feb 7 '11 at 17:14
  • 1
    $\begingroup$ Opps! I forgot my manners. Thanks for the help! $\endgroup$ Feb 7 '11 at 17:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.