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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.

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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 $z=-1$. –  Wadim Zudilin Feb 7 '11 at 12:42
    
@Wadim: I am not convinced either. I did the numerics on the unit circle and got the same. The value attains its minimum at $z=-1.$ –  Daniel Parry Feb 7 '11 at 13:27
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@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... –  Wadim Zudilin Feb 7 '11 at 13:46

1 Answer 1

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):

\[{\def\Li{\text{Li}}}\Re\left[\frac{3\Li_1(z)\Li_3(z)-2\Li_2(z)\Li_2(z)}{\Li_2(z)\Li_3(z)}\right]\]

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 :-)

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

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