Using the concepts of Summability Calculus but generalized such that the lower bound for sums and products is also variable, we can prove that the solution to the following PDE: $$P^2\frac{\partial^2P}{\partial x\partial y}=(P^2+1)\frac{\partial P}{\partial x}\frac{\partial P}{\partial...
No. It's ##\ln(n)+\gamma## (where ##\gamma## is the Euler-Mascheroni constant). However, ##\ln n## is pretty close to the harmonic numbers so replacing the numbers with that function shouldn't change anything.
The limit is $$\lim_{n\rightarrow\infty}\left(\frac{H_n}{n^2}\right)^n=\lim_{n\rightarrow\infty}\frac{H^n_n}{n^{2n}}$$Where ##H_n## is the ##n##th harmonic number. The numerator grows much slower than the denominator and so the answer is just ##0##.
After taking in all your feedback, I submitted a new version of my paper, where instead the main result is called the simple result and is defined as a lemma, not a theorem. I also solved the "conjecture" in my paper and replaced it with a theorem, deriving an asymptotic for the roots of the...
There's a small typo in the OP but I couldn't change it. The reflection formula should be $$\bar{H}_x-\bar{H}_{2-x}=\pi\cot(\pi x)+\left(\frac1{2-x}-\frac1{1-x}-\frac1x\right)\cos(\pi x)$$But this doesn't change the proof of the conjecture.
I solved the conjecture. Take a look at the function ##\bar{H}_x-\bar{H}_{2-x}+\ln2##, which is just the RHS of the reflection formula added by ##\ln2##. As ##x\rightarrow-\infty## we have that ##\bar{H}_x-\bar{H}_{2-x}+\ln2\sim\bar{H}_x## and the RHS approaches ##\ln2+\pi\cot(\pi x)##. So we...