Product of sums which equal to sum of product

Question

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?

Answer 1

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.

Answer 2

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).

Answer 3

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.