Product of sums which equal to sum of product We can be sure that
$$\left(\sum\limits_{k=0}^{n}\frac{1}{k+1}\right)\left(\sum\limits_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{k+1}\right)= \sum\limits_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{(k+1)^2}$$
Is there any similar identities or some types of generalization to find them?
 A: A computer  search will produce the  identity for $p\ge 1$  a positive
integer:
$$\bbox[5px,border:2px solid #00A000]{
\sum_{k=0}^n \frac{1}{k+p}
\sum_{k=0}^n {n\choose k} \frac{(-1)^k}{k+p}
= \sum_{k=0}^n {n\choose k} \frac{(-1)^k}{(k+p)^2}.}$$
This is
$$(H_{n+p} - H_{p-1})
\sum_{k=0}^n {n\choose k} \frac{(-1)^k}{k+p}
= \sum_{k=0}^n {n\choose k} \frac{(-1)^k}{(k+p)^2}.$$
To evaluate the sum on the LHS we introduce
$$f(z) = (-1)^n \frac{n!}{z+p} \prod_{q=0}^n \frac{1}{z-q}.$$
We then obtain
$$\mathrm{Res}_{z=k} f(z)
= (-1)^n \frac{n!}{k+p} 
\prod_{q=0}^{k-1} \frac{1}{k-q} \prod_{q=k+1}^n \frac{1}{k-q}
\\ = (-1)^n \frac{n!}{k+p} \frac{1}{k!} \frac{(-1)^{n-k}}{(n-k)!}
\\ = (-1)^k \frac{1}{k+p} {n\choose k}.$$
With residues adding to zero we find
$$\sum_{k=0}^n {n\choose k} \frac{(-1)^k}{k+p}
= - \mathrm{Res}_{z=-p} f(z) - \mathrm{Res}_{z=\infty} f(z).$$
Now for the residue at infinity we get formally that
$$\mathrm{Res}_{z=\infty} f(z) =
-  \mathrm{Res}_{z=0} \frac{1}{z^2} f(1/z)
\\ = (-1)^{n+1} n! \times \mathrm{Res}_{z=0} \frac{1}{z^2}
\frac{1}{1/z+p} \prod_{q=0}^n \frac{1}{1/z-q}
\\ = (-1)^{n+1} n! \times \mathrm{Res}_{z=0} \frac{1}{z^2}
\frac{z}{1+pz} \prod_{q=0}^n \frac{z}{1-qz}
\\ = (-1)^{n+1} n! \times \mathrm{Res}_{z=0} z^n
\frac{1}{1+pz} \prod_{q=0}^n \frac{1}{1-qz} = 0.$$
We also have
$$- \mathrm{Res}_{z=-p} f(z) =
(-1)^{n+1} n! \prod_{q=0}^n \frac{1}{-p-q}
\\ =  n! \prod_{q=0}^n \frac{1}{p+q}
= n! \frac{(p-1)!}{(n+p)!} = \frac{1}{p} {n+p\choose p}^{-1}.$$
We thus obtain for the LHS the closed form
$$\frac{1}{p} {n+p\choose p}^{-1} (H_{n+p} - H_{p-1}).$$
We use 
$$g(z) = (-1)^n \frac{n!}{(z+p)^2} \prod_{q=0}^n \frac{1}{z-q}.$$
for the  RHS. Observe that the  residue at infinity is  certainly zero
because we are  dividing the bound on $|z|=R$ of  $|f(z)|$ by an extra
factor $R-p.$ This leaves the residue at $z=-p$ and we get
$$- \mathrm{Res}_{z=-p} g(z) =
(-1)^{n+1} n! 
\left.\left(\prod_{q=0}^n \frac{1}{z-q}\right)'\right|_{z=-p}
\\ = (-1)^{n} n! 
\left.\prod_{q=0}^n \frac{1}{z-q}
\sum_{q=0}^n \frac{1}{z-q}
\right|_{z=-p}
\\ = (-1)^{n+1} n! 
\left.\prod_{q=0}^n \frac{1}{z-q}
\right|_{z=-p}
\sum_{q=0}^n \frac{1}{p+q}
\\ = \frac{1}{p} {n+p\choose p}^{-1} (H_{n+p} - H_{p-1}).$$
This  concludes  the proof.   With  this  curious identity  the  first
impression  is  that  we  have  not  understood  the  rules  of  basic
arithmetic involving  sums and products,  yet it reveals itself  to be
true.
A: For anyone who is curious, here is a proof of the above equation. The general technique is to represent these sums as integrals of particular functions.
$$\sum_{k=0}^n\frac1{k+1}=\int_0^1\sum_{k=0}^nx^k\,dx=\int_0^1\frac{1-x^{n+1}}{1-x}\,dx\tag1$$
$$\sum_{k=0}^n\binom{n}k\frac{(-1)^k}{k+1}=\int_0^1 (1-x)^n\,dx=\int_0^1x^n\,dx=\frac1{n+1}\tag2$$
The last summation is the trickiest to simplify. The presence of $1/(k+1)^2$ implies there are two integrations. First, note that
$$
\int_0^x (1-t)^n\,dt = \sum_{k=0}^n\binom{n}k (-1)^k\frac{x^{k+1}}{k+1}
$$
Therefore,
$$
\int_0^1\frac1x\int_0^x (1-t)^n\,dt\,dx = \sum_{k=0}^n\binom{n}k (-1)^k\frac{1}{(k+1)^2}
$$
The left hand integral is
$$
\begin{align}
\int_0^1\int_0^x \frac1x(1-t)^n\,dt
&=\int_0^1\int_{t}^1\frac1x(1-t)^n\,dx\,dt
\\&=\int_0^1(1-t)^n(-\log t)\,dt
\\&=-\int_0^1t^n\log({1-t})\,dt
\end{align}
$$
Noting that $d(\frac{1-t^{n+1}}{n+1})=-t^{n}\,dt$, and integrating by parts,
$$
\begin{align}
\sum_{k=0}^n\binom{n}k (-1)^k\frac{1}{(k+1)^2}
&=\int_0^1 \log(1-t)\,d\left(\frac{1-t^{n+1}}{n+1}\right)
\\&=\require{cancel}\cancelto{0}{\Big(\log(1-t)\frac{1-t^{n+1}}{n+1}\Big)\Bigg|_0^1}-\int_0^1 \frac{1-t^{n+1}}{n+1}\,d\log(1-t)
\\&=\frac1{n+1}\int_0^1 \frac{1-t^{n+1}}{1-t}\,dt\tag3
\end{align}
$$
It is now easy to see that (1) times (2) equals (3).
A: Let $e_k=1$. You want sequences $a_k,\,b_k$ with $\sum_k a_k \sum_k b_k =\sum_k a_k b_k$, or in terms of inner products $a\cdot e \, e\cdot b = a\cdot b$. This equation would be easy to satisfy for $3$-dimensional  vectors: choose your favourite vector $c$ and take $a=c\times e,\, b=a\times c$ so $a\cdot e =a\cdot b =0$. And now you can stitch together infinitely many triples for the infinite-sequence problem, scaling the triples as you go so the resulting sums are convergent. So there are clearly a lot of solutions.
