Prove $lim_{nmapsto 0}[(psi(n)+gamma)psi^{(1)}(n)-frac12psi^{(2)}(n)]=2zeta(3)$

We shall be using the following reflection formulas to evalute the limit.

$$psi_0(1-x)-psi(x)=frac{pi}{tan pi x}\ psi_1(1-x)+psi_1(x)=frac{pi^2}{sin^2 pi x}\psi_2(1-x)-psi_2(x)=frac{d^2}{dx^2}(picotpi x)=frac{2pi^3cotpi x}{sin^2pi x}$$

These give us $$lim_{xto 0}left[left(psi(1-x)+gamma -frac{pi}{tan pi x}right)left(frac{pi^2}{sin^2pi x}-psi_1(1-x)right)-frac{1}{2}left(psi_2(1-x)-frac{2pi^3cotpi x}{sin^2pi x}right)right]$$ Making the use of $displaystyle lim_{xto 0} x^{-1}sin x =1=lim_{xto 0}x^{-1} tan x$ the last expression can be reduced to $$lim_{xto 0}left[left(psi(1-x)+gamma -frac{1}{ x}right)left(frac{1}{ x^2}-psi_1(1-x)right)-frac{1}{2}left(psi_2(1-x)-frac{2}{x^3}right)right] \=-frac{1}{2}psi_2(1-x)+lim_{xto 0} left(frac{psi(1-x)+gamma +xpsi_1(1-x)}{x^2}right)$$ since the latter limit obtained attains $0/0$ form so we evaluate it by L-hopital's rule $$-frac{1}{2}psi_2(1-x)+lim_{xto 0}frac{-psi_1(1-x)+0+psi_1(1-x)-xpsi_2(1-x)}{2x}=lim_{xto 0}left(-frac{1}{2}psi_2(1-x)-frac{1}{2}psi_2(1-x)right)=-psi_2(1)= -(-1)^{3} 2!zeta(3,1)=2zeta(3)$$

Solution due to my friend Khalef Ruhemi ( he is not a MSE user ):

Let $f(x)=(psi(x)+gamma)psi^{(1)}(x)-frac12psi^{(2)}(x)$

Since

$$psi(x)=psi(x+1)-frac1x$$

$$Longrightarrow psi^{(1)}(x)=psi^{(1)}(x+1)+frac1{x^2}$$

$$Longrightarrow psi^{(2)}(x)=psi^{(2)}(x+1)-frac{2}{x^3}$$

we have

$$f(x)=(gamma+psi(x+1))psi^{(1)}(x+1)+frac{gamma+psi(x+1)-frac12x^2psi^{(2)}(x+1)-xpsi^{(1)}(x+1)}{x^2}$$

Thus,

$$lim_{xto 0}f(x)=$$ $$underbrace{lim_{xto 0}(gamma+psi(x+1))psi^{(1)}(x+1)}_{0 text{as $psi(1)=-gamma$}}+underbrace{lim_{xto 0}frac{gamma+psi(x+1)-frac12x^2psi^{(2)}(x+1)-xpsi^{(1)}(x+1)}{x^2}}_{text{L'Hopital}}$$

$$=lim_{xto 0}frac{psi^{(1)}(x+1)-xpsi^{(2)}(x+1)-frac12 x^2psi^{(3)}(x+1)-psi^{(1)}(x+1)-xpsi^{(2)}(x+1)}{2x}$$

$$=lim_{xto 0}(-psi^{(2)}(x+1)-frac14 xpsi^{(3)}(x+1))=-psi^{(2)}(1)=2zeta(3)$$