How to write or input the symbol $S_k(N,x)$ inside the summation $\sum_{n=1}^{N-1} \binom{2n}{n} x^n \sum_{u=0}^{l+1} \binom{l+1}{u}n^u$

Question

Suppose $S_k(N,x)=\sum_{n=1}^{N-1} \binom{2n}{n} n^{k+1}x^n$.

Now I want to write or input the symbol $S_k(N,x)$ inside the summation $\sum_{n=1}^{N-1} \binom{2n}{n} x^n \sum_{u=0}^{l+1} \binom{l+1}{u}n^u, \ \cdots \cdots (1)$.

I tried in the following way

$\sum_{n=1}^{N-1} \binom{2n}{n} x^n \sum_{u=0}^{l+1} \binom{l+1}{u}n^u \\ =\sum_{n=1}^{N-1} \binom{2n}{n} \frac{x^n}{n} \sum_{u=0}^{l+1} \binom{l+1}{u}n^{u+1} \\ =\left(\sum_{n=1}^{N-1} \frac{1}{n} \right) \left( \sum_{n=1}^{N-1} \binom{2n}{n} x^n \sum_{u=0}^{l+1} \binom{l+1}{u}n^{u+1} \right), \ \text{but this splitting it is not true} \\ =\left(\sum_{n=1}^{N-1} \frac{1}{n} \right) \left(\sum_{u=0}^{l+1} \binom{l+1}{u} S_u(N,x) \right) \\ =H_n*\left(\sum_{u=0}^{l+1} \binom{l+1}{u} S_u(N,x) \right), \ \text{where $H_n$ is harmonic number}$.

Thus since the splitting is not true, my method is wrong.

Can you help in any other way to the term $S_k(N,x)$ into the expression $(1)$ ?

Thanks in advance

Answer 1

The following might be helpful. We have \begin{align*} \color{blue}{S_k(N,x)}=\sum_{n=1}^{N-1}\binom{2n}{n}n^{k+1}x^n \end{align*}

We obtain \begin{align*} \sum_{n=1}^{N-1}&\binom{2n}{n}x^n\sum_{k=0}^{l+1}\binom{l+1}{k}n^k\\ &=\sum_{k=0}^{l+1}\binom{l+1}{k}\sum_{n=1}^{N-1}\binom{2n}{n}n^kx^n\tag{1}\\ &=\sum_{k=0}^{l+1}\binom{l+1}{k}\color{blue}{S_{k-1}(N,x)} \end{align*}

Comment:

• In (1) we rearrange the sums to easier substitute $S_k(N,x)$.

Supplement: A relationship between $S_k(N,x)$ and $S_{k-1}(N,x)$

We have \begin{align*} \int_0^x\frac{1}{t}\color{blue}{S_k(N,t)}\,dt&=\int_{0}^x\sum_{n=1}^{N-1}\binom{2n}{n}n^{k+1}t^{n-1}\,dt\\ &=\sum_{n=1}^{N-1}\binom{2n}{n}n^{k+1}\int_{0}^xt^{n-1}\,dt\\ &=\sum_{n=1}^{N-1}\binom{2n}{n}n^kx^n\\ &=\color{blue}{S_{k-1}(N,x)}\tag{2} \end{align*}

Combining (1) and (2) we can write \begin{align*} \sum_{n=1}^{N-1}&\binom{2n}{n}x^n\sum_{k=0}^{l+1}\binom{l+1}{k}n^k\\ &=\sum_{k=0}^{l+1}\binom{l+1}{k}S_{k-1}(N,x)\\ &=\int_0^x\frac{1}{t}\sum_{k=0}^{l+1}\binom{l+1}{k}\color{blue}{S_k(N,t)}\,dt \end{align*}