@BalarkaSen I did the comparison in Mathematica:
Do[
Print[{N[{Sum[
Binomial[2 (n - 1), (n - 1)]/4^(n - 1)/n^k, {n, 1, Infinity}] -
HypergeometricPFQ[Flatten[{{1/2}, Table[1, {n, 1, k}]}],
Table[2, {n, 1, k}], 1]}],
{HypergeometricPFQ[Flatten[{{1}, Table[1, {n, 1, k}]}],
Table[2, {n, 1, k}], 1]}}], {k, 1, 4}]
After some tests I think that
Conjecture 1 Let $x \in \mathbb{R}_+$ then
$$
\int_{0}^{x}\left\{\,1 \over t\,\right\}\,{\rm d}t
= 1 - \gamma + H_{\left\{1/x\right\}} - x\left\lfloor\, 1/x\,\right\rfloor
+ \log\left(\,x\,\right)
$$
Where $\{x\} = x - \left\lfloor\, x\,\right\rfloor$ is the...
Question: $xu_x-xyu_y-u=0$ for all $ (x,y)$
My attempt: Our characteristic curve is in the form of $\frac{dy}{dx}$. Since our $dy = -xy$ and $dx = x$ we have the following separable equation .
$\frac{dy}{dx} = \frac{-xy}{x}$
so that leaves us with
$\frac{dy}{dx} = -y$
$\frac{-1}{y}dy = dx$
...
$$\frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\cdots=\frac{1}{2}$$
Rewrite the sum as
\begin{align}
\frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\cdots
&=\sum^\infty_{n=0}\frac{(2n+1)!!}{(2n+4)!!}\\
&=\sum^\infty_{n=0}\frac{(2n+1)!}{2^{2n+2}n!(n+2)!}\\
&=\sum^\infty_{n=0}\frac{1}{2n+2...
