I need help to prove that :$\sum_{k=1}^{n}\frac{2^k}{k}\frac{(2n-k)!n!}{(2n)!(n-k)!}=\sum_{k=1}^n\frac{1}{2k-1}$

Question

I need help to prove that : $$ \sum_{k = 1}^{n}\frac{2^{k}}{k} \frac{\left(2n-k\right)!\,n!} {\left(2n\right)!\,\left(n - k\right)!} = \sum_{k = 1}^{n}\frac{1}{2k - 1} $$

I tried all the tricks I know, but they don't work. It took me a long time trying to find the solution (for more than 4 hours).

My idea Let $$S_n=\sum_{k=1}^{n}\frac{2^k}{k}\frac{(2n-k)!n!}{(2n)!(n-k)!}$$ It suffices to prove that $S_{n+1}-S_n=\frac{1}{2n+1}$

I tried to prove this as well, and it started to get more and more complicated.

Answer 1

We seek to show that

$${2n\choose n}^{-1} \sum_{k=1}^n \frac{2^k}{k} {2n-k\choose n-k} = \sum_{k=1}^n \frac{1}{2k-1}.$$

Working with the binomial coefficient in the sum on the LHS we get

$$\frac{(2n)^{\underline n}}{(2n)^{\underline k}} \frac{1}{(n-k)!} = {2n\choose n} {n\choose k} {2n\choose k}^{-1}$$

and hence we have

$$\sum_{k=1}^n {n\choose k} \frac{2^k}{k} {2n\choose k}^{-1}.$$

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

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

Apply to our sum to get

$$[v^{2n}] \log\frac{1}{1-v} (v-1)^{2n} \sum_{k=1}^n {n\choose k} 2^k (v-1)^{-k}.$$

Here we get two pieces, the first is

$$- [v^{2n}] \log\frac{1}{1-v} (v-1)^{2n} = - \sum_{q=1}^{2n} \frac{(-1)^q}{q} {2n\choose q}.$$

and the second

$$[v^{2n}] \log\frac{1}{1-v} (v-1)^{2n} \left[1+\frac{2}{v-1}\right]^n \\ = [v^{2n}] \log\frac{1}{1-v} (v-1)^{n} (v+1)^n = \sum_{q=1}^n \frac{(-1)^q}{2q} {n\choose q}.$$

Sum with two instances appearing

We clearly require

$$\sum_{q=1}^m \frac{(-1)^q}{q} {m\choose q}.$$

With this in mind we introduce

$$f(z) = (-1)^m m! \frac{1}{z} \prod_{r=0}^m \frac{1}{z-r}.$$

which has the property that for $1\le q\le m$

$$\;\underset{z=q}{\mathrm{res}}\; f(z) = (-1)^m m! \frac{1}{q} \prod_{r=0}^{q-1} \frac{1}{q-r} \prod_{r=q+1}^m \frac{1}{q-r} \\ = (-1)^m m! \frac{1}{q} \frac{1}{q!} \frac{(-1)^{m-q}}{(m-q)!} = \frac{(-1)^q}{q} {m\choose q}.$$

With the residue at infinity being zero we have that our desired sum is contribution from minus the residue at zero, which gives

$$(-1)^m m! \left. \prod_{r=1}^m \frac{1}{z-r} \sum_{r=1}^m \frac{1}{r-z} \right|_{z=0} = -H_m.$$

Collecting the pieces

Apply to the two pieces to get

$$\frac{1}{2} (-H_n) + H_{2n} = - \sum_{q=1}^n \frac{1}{2q} + \sum_{q=1}^{2n} \frac{1}{q} = \sum_{k=1}^n \frac{1}{2k-1}.$$

This is the claim and we may conclude.

Answer 2

A partial answer

$$\begin{gather*} S_{n+1} - S_n = \sum_{k=1}^{n+1}\frac{2^k}{k}\frac{(2n+2-k)!(n+1)!}{(2n+2)!(n-k+1)!} - \sum_{k=1}^{n}\frac{2^k}{k}\frac{(2n-k)!n!}{(2n)!(n-k)!} \\ = \frac{2^{n+1}}{n+1}\frac{(2n+2-n-1)!(n+1)!}{(2n+2)!(n-n-1+1)!} + \sum_{k=1}^{n}\frac{2^k}{k}\frac{(2n-k)!n!}{(2n)!(n-k)!} - \sum_{k=1}^{n}\frac{2^k}{k}\frac{(2n-k)!n!}{(2n)!(n-k)!} \\ = \frac{2^{n+1}}{n+1}\frac{(n+1)!(n+1)!}{(2n+2)!} \end{gather*} $$ Now we will split this into two cases: one for $n$ odd, and one for $n$ even.

Case 1: $n = 2j-1, \ \ \ j \in \mathbb{N}$

$$\begin{gather*} \frac{2^{n+1}}{n+1}\frac{(n+1)!(n+1)!}{(2n+2)!} \\ = \frac{2^{2j}}{2j}\frac{(2j)!(2j)!}{(4j)!} \\ = \frac{(4j)!!}{(4j)!} \cdot (2j-1)! \\ = \frac{(2j-1)!}{(4j-1)!!} \\ \\ = \frac{(2j-1)!}{\frac{(4j-1)!}{2^{2j-1} (2j-1)!}} \end{gather*}$$ Sadly I am unable to simplify further, if someone sees something, please do edit this answer and add it! I tried using the product definitions, but in my rough I couldn't come up with anything :|

Case 2: $n = 2j, \ \ \ j \in \mathbb{N}$

$$\begin{gather*} \frac{2^{n+1}}{n+1}\frac{(n+1)!(n+1)!}{(2n+2)!} \\ = \frac{2^{2j+1}}{2j+1}\frac{(2j+1)!(2j+1)!}{(4j+2)!} \\ = 2 \cdot (4j)!! \cdot \frac{(2j+1)!}{(4j+2)!} \end{gather*}$$

This case is tougher. I can't think of anything that might simplify this either.