How to evaluate the finite summation of number series

Question

Let $\{a_k\}$ be a number sequence with $a_0=1$ and $$a_k=\frac{\delta^k}{k!}-\frac{\delta^{k-1}}{(k-1)!}$$ when $k\geq 1$.

How to evaluate the following summation? $$ \sum_{k=0}^ra_kk!{(\mu r)}^{-k} \binom{r}{k} $$

Answer 1

We have $a_kk!=\delta^k-\delta^{k-1}k$, thus $$\sum_{k=0}^r a_k k!(\mu r)^{-k}\binom{r}{k}=\sum_{k=0}^r (\frac{\delta}{\mu r})^k\binom{r}{k} -\frac{1}{\delta} \sum_{k=0}^r (\frac{\delta}{\mu r})^k k\binom{r}{k}\\ =\left(1+\frac{\delta}{\mu r} \right)^r-\frac{1}{\delta}\sum_{k=0}^r (\frac{\delta}{\mu r})^k k\binom{r}{k} $$ Notice that from $(1+x)^r=\sum_{k=0}^r \binom{r}{k}x^k$. Differentiating both sides gives $$r(1+x)^{r-1}=\sum_{k=0}^r \binom{r}{k}kx^{k-1}=\frac{1}{x}\sum_{k=0}^r \binom{r}{k}kx^{k}$$ Therefore, $$\sum_{k=0}^r a_k k!(\mu r)^{-k}\binom{r}{k}=\left(1+\frac{\delta}{\mu r} \right)^r-\frac{1}{\delta}\sum_{k=0}^r (\frac{\delta}{\mu r})^k k\binom{r}{k} \\ \left(1+\frac{\delta}{\mu r} \right)^r-\frac{1}{\delta}r(\frac{\delta}{\mu r})\left( 1+\frac{\delta}{\mu r}\right)^{r-1}=\\ \left( 1+\frac{\delta}{\mu r}\right)^{r-1}(1+\frac{\delta}{\mu r}-\frac{1}{\mu})$$