MHB Generating function for trigamma^2

[alyafey22]

In this thread we are looking at the following generating function

$$\sum_{n=1}^\infty [\psi_1(n)]^2 y^n$$

We know that this is as hard as evaluating

$$\sum_{n=1}^\infty [H_n^{(2)}]^2 y^n$$

This is not a tutorial as I have no idea how to solve for a general formula. I'll keep posting my attempts on it. As always all other attempts and suggestions are welcomed.

 

[alyafey22]

Start by the following

$$\int^1_0 x^{k-1} \log(x) \,dx = \frac{1}{k^2}$$

Apply $\sum_k^n$ to both sides and rearrange

$$\int^1_0 \frac{x^{n}}{x-1}\log(x) \,dx = \zeta(2)-H_{n}^{(2)}$$

Multiply by $H_{n}^{(2)}$

$$\int^1_0 \frac{x^{n}H_{n}^{(2)}}{x-1}\log(x) \,dx = H_{n}^{(2)}\zeta(2)-[H_{n}^{(2)}]^2$$

Introduce a new variable $y$

$$\int^1_0 \frac{(yx)^{n}H_{n}^{(2)}}{x-1}\log(x) \,dx = \zeta(2)H_{n}^{(2)}y^n-[H_{n}^{(2)}]^2 y^n$$

Sum both sides

$$\int^1_0 \frac{\mathrm{Li}_2(xy)}{(1-x)(1-xy)}\log(x) \,dx = \sum_{n=1}^\infty [H_{n}^{(2)}]^2 y^n-\zeta(2)\frac{\mathrm{Li}_2(y)}{1-y}$$

$$\sum_{n=1}^\infty [H_{n}^{(2)}]^2 y^n=\int^1_0 \frac{\mathrm{Li}_2(xy)}{(1-x)(1-xy)}\log(x) \,dx +\zeta(2)\frac{\mathrm{Li}_2(y)}{1-y}$$

 

[fredoniahead]

Cool thread, Z.

Here are a few thoughts. I am going to use $$\psi_{1}(n+1)$$ simply for ease of computations. We can adjust it for $\psi_{1}(n)$.

Attempt to find a gen. func. for $$\sum_{n=1}^{\infty}\left(\psi_{1}(n+1)\right)^{2}y^{n}$$

Note that:

$$\psi_{1}(n+1)=\zeta(2)-H_{n}^{(2)}$$

Square this:

$$\left(\zeta(2)-H_{n}^{(2)}\right)^{2}y^{n}$$

$$=\left(\frac{\pi^{4}}{36}-\frac{\pi^{2}}{3}H_{n}^{(2)}+\left(H_{n}^{(2)}\right)^{2}\right)y^{n}$$

Z, you already derived the $\sum_{n=1}^{\infty}\left(H_{n}^{(2)}\right)^{2}y^{n}$ part.

So, we need:

$$\frac{\pi^{4}}{36}\sum_{n=1}^{\infty}y^{n}-\frac{\pi^{2}}{3}\sum_{n=1}^{\infty}H_{n}^{(2)}y^{n}$$

$$=\frac{\pi^{4}y}{36(1-y)}-\frac{\pi^{2}}{3}\left(\frac{-2Li_{2}\left(\frac{y}{y-1}\right)}{1-y}\right)$$To instead obtain:

$$\sum_{n=1}^{\infty}\left(\psi_{1}(n)\right)^{2}y^{n}$$ as originally posted, one may use

$$\psi_{1}(n)=\zeta(2)-H_{n-1}^{(2)}=\zeta(2)-\left(H_{n}^{(2)}-\frac{1}{n^{2}}\right)$$

Then, proceed as before.

 

[fredoniahead]

Hey Z:

Here is another perhaps we can find a gen. func for. I feel I should already have this one.

$$\sum_{n=1}^{\infty}(H_{n})^{3}y^{n}$$

I derived several fun sums by successive diffing of the digamma.

If we note: $$G(x)=\frac{\Gamma(x)n!}{\Gamma(n+x+1)}$$

and $$P(x)=\psi(x)-\psi(x+n+1)$$

then, $$G'(x)=G(x)P(x), \;\ G''(x)=G(x)P^{2}(x)+G(x)P'(x), \;\ P^{(m)}(1)=(-1)^{m}m!H_{n+1}^{m+1}$$

$$\psi^{(m)}(1)=(-1)^{m+1}m!\zeta(m+1)$$

this finally leads to, for instance,

$$24\zeta(5)=2\sum_{n=1}^{\infty}\frac{H_{n}^{(3)}}{n^{2}}+3\sum_{n=1}^{\infty}\frac{H_{n}^{(2)}H_{n}}{n^{2}}+\sum_{n=1}^{\infty}\frac{\left[H_{n}\right]^{3}}{n^{2}}$$As an aside, there is also Morley's identity. Playing around (say, diffing w.r.t 'a' ) with this can lead to other fun sums and so forth:

$$\sum_{n=1}^{\infty}\left[\frac{\Gamma(a+n)}{n!\Gamma(a)}\right]^{3}=\cos\left(\frac{\pi a}{2}\right)\cdot \frac{\Gamma\left (1-\frac{3a}{2}\right)}{\Gamma^{3}\left (1-\frac{a}{2}\right)}-1$$$$

 

[alyafey22]

\
If we note: $$G(x)=\frac{\Gamma(x)n!}{\Gamma(n+x+1)}$$

and $$P(x)=\psi(x)-\psi(x+n+1)$$

then, $$G'(x)=G(x)P(x), \;\ G''(x)=G(x)P^{2}(x)+G(x)P'(x), \;\ P^{(m)}(1)=(-1)^{m}m!H_{n+1}^{m+1}$$

$$\psi^{(m)}(1)=(-1)^{m+1}m!\zeta(m+1)$$

this finally leads to, for instance,

$$24\zeta(5)=2\sum_{n=1}^{\infty}\frac{H_{n}^{(3)}}{n^{2}}+3\sum_{n=1}^{\infty}\frac{H_{n}^{(2)}H_{n}}{n^{2}}+\sum_{n=1}^{\infty}\frac{\left[H_{n}\right]^{3}}{n^{2}}$$

Wow C, nice procedure to find high order Euler sums.

 

[fredoniahead]

Here is another to consider, Z. This converges to $2\zeta(3)-\zeta(2)$

$$\sum_{n=1}^{\infty}H_{n}\left(\zeta(3)-H_{n}^{(3)}\right)=-1/2\sum_{n=1}^{\infty}H_{n}\psi^{(2)}(n+1)$$

 

[alyafey22]

Here is another to consider, Z. This converges to $2\zeta(3)-\zeta(2)$

$$\sum_{n=1}^{\infty}H_{n}\left(\zeta(3)-H_{n}^{(3)}\right)=-1/2\sum_{n=1}^{\infty}H_{n}\psi^{(2)}(n+1)$$

Hey C , I am busy with exams. I will take a look at the new stuff after I finish.

 

[fredoniahead]

Hey Z man:). Here is two other interesting Euler-like sums that do converge. What are your thoughts on finding a closed form?.
$$\sum_{n=1}^{\infty}\left[-(H_{n}^{(2)})^{3}+\frac{\pi^{2}}{2}(H_{n}^{(2)})^{2}-\frac{\pi^{4}}{12}H_{n}^{(2)}+\frac{\pi^{6}}{216}\right]...[1]$$

and/or

$$\sum_{n=1}^{\infty}\frac{\left(2(H_{n}^{(2)})^{2}-\frac{2\pi^{2}}{3}H_{n}^{(2)}+\frac{\pi^{4}}{18}\right)}{n+1}...[2]$$

both of these have closed forms in terms of zeta values...I am pretty sure.

[2] I managed to come up with when using Abel summation to expand and sum [1]:

$$\sum_{n=1}^{\infty}\left(\zeta(2)-H_{n}^{(2})\right)^{3}=\sum_{n=1}^{\infty}\left[-(H_{n}^{(2)})^{3}+\frac{\pi^{2}}{2}(H_{n}^{(2)})^{2}-\frac{\pi^{4}}{12}H_{n}^{(2)}+\frac{\pi^{6}}{216}\right]$$

It is part of the expansion of the Abel sum, $b_{n}-b_{n+1}$ : $$\sum_{n=1}^{\infty}n\left(\psi_{1}^{3}(n+1)-\psi_{1}^{3}(n+2)\right)$$. Well, if you step through it you'll know what I mean.

where $$\psi_{1}(n+1)=\zeta(2)-H_{n}^{(2)}, \;\ \psi_{1}(n+2)=\zeta(2)-H_{n+1}^{(2)}, \;\ H_{n+1}^{(2)}=H_{n}^{(2)}+\frac{1}{(n+1)^{2}}$$

I used the difference of two cubes factorization to write this as several Euler sums. All are known values except [2] above. This is one portion of the expansion that is conditionally convergent.EDIT:

I managed to find the closed form for [2].

It is $$10\zeta(2)\zeta(3)+2\zeta(5)-5\zeta(4)-\frac{4\pi^{2}}{3}\zeta(3)$$

I can outline my method if anyone is interested.