how to prove that $\psi_1(x)=\sum_{n=0}^{\infty}\frac{B_n}{x^{n+1}}$ ?

Question

how to prove that $$\psi_1(x)=\sum_{n=0}^{\infty}\frac{B_n}{x^{n+1}}$$

where $\psi_1(z)$ is Trigamma function

and $B_n$ is Bernoulli number

Answer 1

Let us use $x=e^{-u}$ in the integral definition of the trigamma function to get: $$\psi_1(z)=-\int _{0}^{1}\!{\frac {{x}^{z-1}\ln \left( x \right) }{1-x}}{dx}= \int _{0}^{\infty }\!{\frac {u{{\rm e}^{-uz}}}{1-{{\rm e}^{-u}}}}{du}.\tag{1}$$ Then from the exponential generating function for the Bernoulli polynomials: $$\frac{u{\rm e}^{xu}}{{\rm e}^{u}-1}=\sum _{n=0}^{\infty }{\frac { {\it B_n} \left( x \right) {u}^{n}}{n!}},$$ we write: $$\frac{u}{1-{\rm e}^{-u}}=\sum _{n=0}^{\infty }{\frac { (-1)^n{\it B_n} {u}^{n}}{n!}},$$ wher $B_n=B_n(0)$ is a Bernoulli number of the first kind, and we have: $$\psi_1(z)=\int _{0}^{\infty }\!{\frac {u{{\rm e}^{-uz}}}{1-{{\rm e}^{-u}}}}{du} =\int _{0}^{\infty }\left(\!\sum _{n=0}^{\infty }{\frac { \left( -1 \right) ^{n}{\it B}_n {u}^{n}}{n!}}\right){{\rm e}^{-uz}}{du}.\tag{2}$$ It would be useless to try and justify switching integration and summation in ($2$) as @gammatester has shown the resulting sum does not converge for any $z$ anyway, but the asymptotic expansion ignores this fact to some extent and manipulates things in a purely formal sense, in the hope that truncating the series offers a good approximation. Continuing in this manner then we use: $$\Gamma \left( r+1 \right) ={z}^{r+1}\int _{0}^{\infty }\!{u}^{r}{ {\rm e}^{-uz}}{du},$$ to get: $$\int _{0}^{\infty }\!{u}^{n}{{\rm e}^{-uz}}{du}={\frac {n!}{{z}^{n+1}} },$$ and eventually: $$\psi_1(z)=\int _{0}^{\infty }\!{\frac {u{{\rm e}^{-uz}}}{1-{{\rm e}^{-u}}}}{du} =\sum _{n=0}^{\infty }{\frac { \left( -1 \right) ^{n}{\it B}_n }{{z}^{n+1}}}.\tag{3}$$ If we now introduce the Bernoulli numbers of the second kind, which for convenience we will denote $B'_n$, and note that by definition: $$B'_{2n}=B_{2n},$$ $$B'_{2n+1}=B_{2n+1}=0:2n+1>1,$$ $$B'_{1}=-B_{1}=-\frac{1}{2},$$ we obtain: $$\psi_1(z)=\sum _{n=0}^{\infty }{\frac { \left( -1 \right) ^{n}{\it B}_n }{{z}^{n+1}}}=\sum _{n=0}^{\infty }{\frac { {\it B}_n' }{{z}^{n+1}}}.\tag{4}$$

Answer 2

To give a formal answer: The sum is not convergent for any $x$. Proof: We have $$\psi_1(x)=\sum_{n=0}^{\infty}\frac{B_n}{x^{n+1}}= \frac{1}{x} + \frac{1}{2x^2} + \frac{1}{x}\sum_{n=1}^{\infty}\frac{B_{2n}}{x^{2n}}$$ Using the formula http://dlmf.nist.gov/24.9#E6 for the Bernoulli numbers, the absolute value of the general term of the last sum is $$|\frac{B_{2n}}{x^{2n}}| > 4 \sqrt{\pi n}\frac{(\frac{n}{\pi e})^{2n}}{x^{2n}} = 4 \sqrt{\pi n}(\frac{n}{\pi e x})^{2n},$$ which shows that the sum does not converge.