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Prove that: $$\alpha_{n}=\sum_{k=0}^{n} \binom{n+k}{n-k}\beta_{k} \Leftrightarrow \beta_{n}=\sum_{k=0}^{n} (-1)^{n-k}\frac{2k+1}{2n+1} \binom{2n+1}{n-k}\alpha_{k}$$ What I have tried:

$\Longrightarrow ):$ $$\begin{align*} \sum_{k=0}^{n} (-1)^{n-k} \frac{2k+1}{2n+1}\binom{2n+1}{n-k}\alpha_{k} &=\sum_{k=0}^{n} (-1)^{n-k} \frac{2k+1}{2n+1}\binom{2n+1}{n-k} \sum_{j=0}^{k} \binom{k+j}{k-j}\beta_{j} \\ &=\sum_{k=0}^{n} \sum_{j=0}^{k} (-1)^{n-k} \frac{2k+1}{2n+1}\binom{2n+1}{n-k} \binom{k+j}{k-j}\beta_{j} \\ &=\sum_{j=0}^{n} \sum_{k=j}^{n} (-1)^{n-k} \frac{2k+1}{2n+1}\binom{2n+1}{n-k} \binom{k+j}{k-j}\beta_{j} \\ &=\sum_{j=0}^{n} \beta_{j} \sum_{k=j}^{n} (-1)^{n-k} \frac{2k+1}{2n+1}\binom{2n+1}{n-k} \binom{k+j}{k-j} \end{align*}$$

I have verified that $RHS=\beta_{n}$ when $j=n$ and $RHS=0$ when $j=n-1$, so I guess that $$\sum_{k=j}^{n} (-1)^{n-k} \frac{2k+1}{2n+1}\binom{2n+1}{n-k} \binom{k+j}{k-j}=0$$ established when $$0\le j< n $$

$\Longleftarrow ):$

$$\begin{align*} \sum_{k=0}^{n}\binom{n+k}{n-k}\beta_{k} &=\sum_{k=0}^{n}\binom{n+k}{n-k} \sum_{j=0}^{k} (-1)^{k-j} \frac{2j+1}{2k+1} \binom{2k+1}{k-j} \alpha_{j} \\ &=\sum_{k=0}^{n} \sum_{j=0}^{k}\binom{n+k}{n-k} (-1)^{k-j} \frac{2j+1}{2k+1} \binom{2k+1}{k-j}\alpha_{j} \\ &=\sum_{j=0}^{n} \sum_{k=j}^{n}\binom{n+k}{n-k} (-1)^{k-j} \frac{2j+1}{2k+1} \binom{2k+1}{k-j}\alpha_{j} \\ &=\sum_{j=0}^{n}\alpha_{j} \sum_{k=j}^{n}\binom{n+k}{n-k} (-1)^{k-j} \frac{2j+1}{2k+1} \binom{2k+1}{k-j} \end{align*}$$ It can verified that$RHS=\alpha_{n}$ when $j=n$ and $RHS=0$ when $j=n-1$,similarly guess that $$\sum_{k=j}^{n}\binom{n+k}{n-k} (-1)^{k-j} \frac{2j+1}{2k+1} \binom{2k+1}{k-j}=0$$ established when $$0\le j< n $$

Although the general idea is there, I don't know how to prove the key two equations. I hope MSE can give me some advice, thank you very much!

anyon
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  • I have no time to check, but I guess a method similar to the one using Pascal matrix would work. https://www.wikiwand.com/en/Pascal_matrix – Taladris Oct 15 '22 at 02:25

1 Answers1

2

First indicator

We seek to evaluate where $n\ge j$

$$\sum_{k=j}^n (-1)^{n-k} \frac{2k+1}{2n+1} {2n+1\choose n-k} {k+j\choose k-j}.$$

This is

$$\sum_{k=0}^{n-j} (-1)^k \frac{2n-2k+1}{2n+1} {2n+1\choose k} {n-k+j\choose n-k-j} \\ = [z^{n-j}] (1+z)^{n+j} \sum_{k\ge 0} (-1)^k \frac{2n-2k+1}{2n+1} {2n+1\choose k} z^k (1+z)^{-k}.$$

Here we have extended to infinity due to the coefficient extractor. Continuing we get two pieces, the first is

$$[z^{n-j}] (1+z)^{n+j} \sum_{k\ge 0} (-1)^k {2n+1\choose k} z^k (1+z)^{-k} \\ = [z^{n-j}] (1+z)^{n+j} \left[1-\frac{z}{1+z}\right]^{2n+1} \\ = [z^{n-j}] \frac{1}{(1+z)^{n+1-j}} = (-1)^{n-j} {2n-2j\choose n-j}.$$

The second is

$$- [z^{n-j}] (1+z)^{n+j} \sum_{k\ge 0} (-1)^k \frac{2k}{2n+1} {2n+1\choose k} z^k (1+z)^{-k} \\ = -\frac{2}{2n+1} [z^{n-j}] (1+z)^{n+j} \sum_{k\ge 1} (-1)^k k {2n+1\choose k} z^k (1+z)^{-k} \\ = - 2 [z^{n-j}] (1+z)^{n+j} \sum_{k\ge 1} (-1)^k {2n\choose k-1} z^k (1+z)^{-k} \\ = 2 [z^{n-j}] z (1+z)^{n+j-1} \sum_{k\ge 0} (-1)^k {2n\choose k} z^k (1+z)^{-k}.$$

This will produce zero when $n=j$. Continuing with $n\gt j$,

$$2 [z^{n-j-1}] (1+z)^{n+j-1} \left[1-\frac{z}{1+z}\right]^{2n} \\ = 2 [z^{n-j-1}] \frac{1}{(1+z)^{n+1-j}} = 2 (-1)^{n-j-1} {2n-2j-1\choose n-j}.$$

Collecting the contributions we find for $n=j$

$$(-1)^{n-n} {2n-2n\choose n-n} = 1$$

and for $n\gt j$

$$(-1)^{n-j} {2n-2j\choose n-j} \left[1-2\frac{n-j}{2n-2j}\right] = 0.$$

This is the claim.

Second indicator

We seek to evaluate where $n\ge j$

$$\sum_{k=j}^n {n+k\choose n-k} (-1)^{k-j} \frac{2j+1}{2k+1} {2k+1\choose k-j}.$$

Expanding the sum term we find

$$(2j+1) \frac{(n+k)!}{(n-k)!} (-1)^{k-j} \frac{1}{(k-j)! \times (k+j+1)!} \\ = (2j+1) {n-j\choose k-j} (-1)^{k-j} {n+k\choose n-j} \frac{1}{k+j+1}.$$

We get for the sum

$$(2j+1) \sum_{k=0}^{n-j} {n-j\choose n-k-j} (-1)^{n-k-j} {2n-k\choose n-j} \frac{1}{n+j-k+1} \\ = (2j+1) [z^{n-j}] (1+z)^{2n} \sum_{k=0}^{n-j} {n-j\choose k} (-1)^{n-k-j} \frac{1}{(1+z)^k} \frac{1}{n+j-k+1} \\ = (2j+1) [z^{n-j}] (1+z)^{2n} [w^{n+j+1}] \log\frac{1}{1-w} \left[ \frac{w}{1+z} -1 \right]^{n-j} \\ = (2j+1) [z^{n-j}] (1+z)^{n+j} [w^{n+j+1}] \log\frac{1}{1-w} \left[ w-1-z \right]^{n-j}.$$

For $n=j$ this will produce

$$(2n+1) [z^0] (1+z)^{2n} \frac{1}{2n+1} = 1$$

as required. Continuing with $n\gt j$ and the extractor in $w$,

$$[w^{n+j+1}] \log\frac{1}{1-w} \sum_{q=0}^{n-j} {n-j\choose q} (w-1)^q (-1)^{n-j-q} z^{n-j-q}.$$

Recall from MSE 4316307 the following identity which was proved there: with $1\le k\le n$

$$\frac{1}{k} {n\choose k}^{-1} = [w^n] \log\frac{1}{1-w} (w-1)^{n-k}.$$

so this becomes

$$(2j+1) [z^{n-j}] (1+z)^{n+j} \sum_{q=0}^{n-j} {n-j\choose q} \frac{1}{n+j+1-q} {n+j+1\choose q}^{-1} (-1)^{n-j-q} z^{n-j-q} \\ = (2j+1) \sum_{q=0}^{n-j} {n-j\choose q} \frac{1}{n+j+1-q} {n+j+1\choose q}^{-1} (-1)^{n-j-q} {n+j\choose q}.$$

We have

$${n+j\choose q} {n+j+1\choose q}^{-1} = \frac{(n+j)! \times (n+j+1-q)!}{(n+j-q)! \times (n+j+1)!}$$

and we obtain at last

$$\frac{2j+1}{n+j+1} \sum_{q=0}^{n-j} {n-j\choose q} (-1)^{n-j-q} = [[n=j]].$$

Once more we have the claim. It appears that the special case $n=j$ was subsumed by the above computation.

Marko Riedel
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