A convergent series for the Trigamma function $\psi_1(n) =\sum_{k=n}^{\infty} \frac1{k^2} $

Question

I just came up with the following convergent series for the Trigamma function defined by

$\psi_1(n) =\sum_{k=n}^{\infty} \frac1{k^2} $.

\begin{align*} \psi_1(n) &=\lim_{m \to \infty} \sum_{j=1}^m \frac{(j-1)!}{j\prod_{i=0}^{j-1}(n+i)}\\ &=\frac1{n}+\frac{1}{2n(n+1)}+\frac{2}{3n(n+1)(n+2)}+\ldots\\ \end{align*}

This contrasts with the usual asymptotic series for $\psi_1(n)$ which is asymptotic, does not converge, and involves the Bernoulli numbers.

I'm sure this isn't new, but I could not find it here.

So, my questions are, as is often the case,

(1) Is this new?

(2) Is there a reasonably simple proof? (Mine is moderately messy.)

I'll post my proof in a few days if anyone is interested.

Thanks

Answer 1

Observe that $\prod_{i=0}^{j-1}(n+i)=\Gamma(n+j)/\Gamma(n)$, so that, as claimed, \begin{align*} \sum_{j=1}^\infty\frac{(j-1)!}{j\prod_{i=0}^{j-1}(n+i)} &=\sum_{j=1}^\infty\frac1j\mathrm{B}(n,j)\\ &=\sum_{j=1}^\infty\frac1j\int_0^1 t^{n-1}(1-t)^{j-1}\,dt\\ &=\int_0^1 t^{n-1}\frac{-\log t}{1-t}\,dt\\ &=\int_0^1 t^{n-1}(-\log t)\sum_{j=0}^\infty t^j\,dt\\ &=\sum_{k=n}^\infty\int_0^1 t^{k-1}(-\log t)\,dt=\sum_{k=n}^\infty\frac1{k^2}. \end{align*}

Answer 2

This is not a proof, not an answer but it is too long for comments.

Let $$a_j=\frac{(j-1)!}{j\prod_{i=0}^{j-1}(n+i)}=\frac{(j-1)!}{j n (n+1)_{j-1}}\qquad \text{and} \qquad S_m=\sum_{j=1}^m a_j$$

$$\frac{(m+1) \Gamma (n+1) \Gamma (m+n+1)}{\Gamma (n) }S_m=$$ $$(m+1) n \psi ^{(1)}(n) \Gamma (m+n+1)-$$ $$\Gamma (m+1) \Gamma (n+1) \, _3F_2(1,m+1,m+1;m+2,m+n+1;1)$$

Simplifying $$S_m=\psi ^{(1)}(n)-\frac{\Gamma (m+1) \Gamma (n)}{(m+1) \Gamma (m+n+1)} \, _3F_2(1,m+1,m+1;m+2,m+n+1;1)$$

What seems interesting is to make a contour plot of the second term; it shows how small it is.

We can also notice that for large values of $m$ $(m\gg n)$, the coefficient $$\frac{\Gamma (m+1) \Gamma (n)}{(m+1) \Gamma (m+n+1)}\sim \frac{ \Gamma (n) }{m^{n+1}}$$

Unfortunately, I have not been to find the asymptotics of the hypergeometric function.