Evaluating an infinite sum involving possibly hypergeometric terms

Question

I was considering the following infinite sum

$$ A(n) = \sum_{k=2}^{\infty}\left[\frac{(-1)^{k+n-1}}{k^n}(0k -1)(k-1)(2k-1)...((n-1)k-1) \right] $$

Some cases:

$$ A(1) = \sum_{k=2}^{\infty}\left[\frac{(-1)^{k+1}}{k} \right]= -\frac{1}{2} + \frac{1}{3} - \frac{1}{4} ... = \ln(2) - 1 $$

A trickier sum to evaluate is: $A(2)$. This resolves to:

$$ A(2) = \sum_{k=2}^{\infty}\left[\frac{(-1)^{k+2}}{k^2} (k-1)\right] = \frac{1}{4} - \frac{2}{9} + \frac{3}{16} ... $$

Some jugglery is required but if we consider

$$ 1 - x + x^2 -x^3 ... = \frac{1}{1+x}$$

And integrate

$$ \ln(1+x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 + ... $$

Divide by x and integrate

$$ \int_{0}^{1} \frac{\ln(1+x)}{x} dx = x - \frac{1}{4}x^2 + \frac{1}{9} x^3 ... $$

Now multiply by (-1), divide by x and differentiate

$$ - \left( \frac{1}{x} \int_{0}^{1} \frac{\ln(1+x)}{x} dx \right)' = \frac{1}{4} - \frac{2}{9} x + \frac{3}{16} x^2 ... $$

Meaning

$$ \left( \frac{1}{x^2} \int_{0}^{1} \frac{\ln(1+x)}{x} dx - \frac{1}{x} \frac{\ln(1+x)}{x} \right)_{@x = 1} = \frac{1}{4} - \frac{2}{9} + \frac{3}{16} ... $$

Through very mysterious techniques of the omniscient Wolfram Alpha this evaluates to

$$ \frac{\pi^2}{12} - \ln(2) = \frac{1}{4} - \frac{2}{9} + \frac{3}{16} ... $$

So now this is totally in the deep end, but I have a suspicion the sum I have given can somehow be related to the riemann zeta function, in some good quality closed form. It's a wild guess but the $\ln(2)$ and now the $\pi^2$ hint me that's a good idea.

How to show this?

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In case it's helpful

I tried to bash this with Sister Celine's method. By writing as

$$ A(n) = \sum_{k=2}^{\infty}\left[\frac{(-1)^{k+n-1}}{k^n}(0k -1)(k-1)(2k-1)...((n-1)k-1) \right] = \sum_{k=2}[ M(n,k) ] $$

And tried to find a recurrence for $M$ but unfortunately after consider $a_0(n) M(n,k) + a_1(n) M(n+1,k) + a_2(n) M(n-1,k) $ it became clear I'm going to need some more terms (possibly a lot more) before I can find a closed recurrence. Unfortunately my maple package has expired (I did send them an email to renew!) so I couldn't throw the might of Shalosh B. Ekhad in the Zeilberger package to close it up.

I suspect though that this would be ample for those techniques.

Answer 1

The trick should be to rewrite (for $n>0$) \begin{align} A(n) &= \sum_{k=2}^{\infty}\frac{(-1)^{k+n-1}}{k^n}(0k -1)(k-1)(2k-1)...((n-1)k-1)\\ &= \sum_{k=2}^{\infty}(-1)^{k-1}\left(\frac 1k\right)\left(\frac 1k-1\right)...\left(\frac 1k-(n-1)\right)\\ &= \sum_{k=2}^{\infty}(-1)^{k-1}\left(\frac 1k\right)_n\\ \end{align}

where $(x)_n$ is the Pochhammer symbol ("falling factorial") and a generating function for the signed Stirling numbers of the first kind (the term $\,i=0\,$ is omitted since $\;s(n,0)=0\,$ for $n>0$) : $$(x)_n=\sum_{i=1}^n s(n,i)\,x^i$$ this gives (for $\,\eta\,$ the Dirichlet eta function or "alternating zeta function") : \begin{align} A(n) &= \sum_{k=2}^{\infty}(-1)^{k-1}\sum_{i=1}^n s(n,i)\,\frac 1{k^i}\\ &= \sum_{i=1}^n s(n,i)\,\sum_{k=2}^{\infty}(-1)^{k-1} \frac 1{k^i}\\ &= \sum_{i=1}^n s(n,i)\,(\eta(i)-1)\quad\text{but}\quad \sum_{i=1}^n s(n,i)=0\quad\text{for}\;n>1\\ &= \sum_{i=1}^n s(n,i)\,\eta(i)\quad\text{for}\;n>1\quad\text{and}\quad s(1,1)(\eta(1)-1)\quad\text{for}\;n=1\\ \\ A(n)&= \begin{cases}s(n,1)\,\log(2)+\sum_{i=2}^n s(n,i)\left(1-2^{1-i}\right)\zeta(i),&n>1\\ \log(2)-1,&n=1\\ \end{cases} \end{align}

Since $\;s(n,1)=(-1)^{n-1}(n-1)!\;$ these expressions are in perfect agreement with the table exposed by Claude Leibovici.

Answer 2

This is not an answer but it is too long for a comment.

Interested by the post, I first wrote (from definition) $$A_n=\sum _{k=2}^{\infty } \frac{(-1)^{k+n} }{k}\left(\frac{k-1}{k}\right)_{n-1}$$ where appears the Pochhammer symbol.

Then, "playing" with a CAS, I obtained the following terms $$A_1=-1+\log (2)$$ $$A_2=\frac{\pi ^2}{12}-\log (2)$$ $$A_3=\frac{3 \zeta (3)}{4}-\frac{\pi ^2}{4}+2 \log (2)$$ $$A_4=-\frac{9 \zeta (3)}{2}+\frac{11 \pi ^2}{12}+\frac{7 \pi ^4}{720}-6 \log (2)$$ $$A_5=\frac{105 \zeta (3)}{4}+\frac{15 \zeta (5)}{16}-\frac{25 \pi ^2}{6}-\frac{7 \pi ^4}{72}+24 \log (2)$$ $$A_6=-\frac{675 \zeta (3)}{4}-\frac{225 \zeta (5)}{16}+\frac{137 \pi ^2}{6}+\frac{119 \pi ^4}{144}+\frac{31 \pi ^6}{30240}-120 \log (2)$$ $$A_7=1218 \zeta (3)+\frac{2625 \zeta (5)}{16}+\frac{63 \zeta (7)}{64}-147 \pi ^2-\frac{343 \pi ^4}{48}-\frac{31 \pi ^6}{1440}+720 \log (2)$$

As you suspect, beside the $(-1)^{n-1}(n-1)!\log(2)$ term, $A_n$ is a linear combination of $\zeta(.)$ functions $$A_n=(-1)^{n-1}(n-1)!\log(2)+\sum_{i=2}^{n}a_i\, \zeta(i)$$

At this point, may I confess that I am totally stuck ?

Edit

Just for clarity, defining $$B_n=A_n-(-1)^{n-1}(n-1)!\log(2)$$ and back to only $\zeta(.)$ functions, we would have $$B_2=\frac 12 \zeta(2)$$ $$B_3=-\frac 32 \zeta(2)+\frac 34 \zeta(3)$$ $$B_4=\frac{11 }{2}\zeta(2)-\frac{9 }{2}\zeta (3)+\frac{7 }{8}\zeta(4)$$ $$B_5=-25 \zeta(2)+\frac{105 }{4}\zeta (3)-\frac{35 }{4}\zeta(4)+\frac{15 }{16}\zeta (5)$$ $$B_6=137 \zeta(2)-\frac{675 }{4}\zeta (3)+\frac{595 }{8}\zeta(4)-\frac{225 }{16}\zeta (5)+\frac{31 }{32}\zeta(6)$$ $$B_7=-882 \zeta(2)+1218 \zeta (3)-\frac{5145 }{8}\zeta(4)+\frac{2625 }{16}\zeta (5)-\frac{651 }{32}\zeta(6)+\frac{63 }{64}\zeta (7)$$

Answer 3

I do not know if what follows is of some help, but: Note that $\prod_{j=0}^{n-1}(jk-1)=P_n(k)$, is a polynomial in $k$. Put $P_n(x)=a_0+a_1x+\cdots a_{n-1}x^{n-1}$, then your sum is $$A(n)=(-1)^{n-1}\sum_{j=0}^{n-1}a_j(\sum_{k\geq 2}\frac{(-1)^k}{k^{n-j}})$$ and $\sum_{k\geq 1}\frac{(-1)^k}{k^{s}}=\eta(s)=(1-2^{1-s})\zeta(s)$ See https://en.wikipedia.org/wiki/Riemann_zeta_function#Generalizations, ( in "the functional equation" part)

The case $j=n-1$ gives you the $\log(2)$ term, up to a constant due to the $k\geq 2$.