Graduate Psi function or digamma function?

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The discussion focuses on the series representation of the psi (digamma) function in the context of Bessel functions. It highlights the expressions for psi functions, specifically psi(m+1) and psi(n+m+1), which correspond to the series sums of 1/k for k from 1 to m and k from 1 to m+n, respectively. The relationship between the psi function and the logarithm of the gamma function is emphasized, with the definition given as ψ(x) = d/dx log(Γ(x)). A heuristic explanation is provided linking the logarithm of the gamma function to the series representation. The conversation concludes with a reference to a collection of formulas related to the gamma and digamma functions for further exploration.
Belgium 12
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Hi everyone,
concerning serie representation of psi function.
In te solution of bessel function of the second kind we have the following expressions for the psi functions

psi(m+1) and psi(n+m+1) then they give the series for the two psi functions ie(or digamma function)

sum
k from 1 to m of 1/k and ( for the first)
sum k from 1 to m+n of 1/k (for the second)

Is it possible to explain where this serie come from maybe using defintion of the psi function.I can't write latex sorry.

Tank you veru much
 
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The LaTeX part is easier than your actual question:
https://www.physicsforums.com/help/latexhelp/

We have ##\psi(x)=\dfrac{d}{dx}\log(\Gamma(x))## and are interested in ##\psi(n)=H_{n-1} -\gamma=H_{n-1}-\lim_{n \to \infty}\left(H_n-\log(n)\right)##.
So we have heuristically a difference quotient of
$$
\log(\Gamma(n))=\log \left(\prod_k k \right)= \sum_k \log k = \sum_k \int_1^k \dfrac{1}{x}\,dx
$$
on the left hand side and the same difference on the right hand side.

An exact proof is a bit more to do. See this nice collection of formulas around ##\Gamma(x)## and ##\psi(x)##:
http://fractional-calculus.com/gamma_digamma.pdf
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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