upper bound for the series $S (x) = \sum_{k=1}^{\infty} \frac{(x_n-n)^k}{k!}$ from $|x_n -(n+1)|\leq x$.

Question

I've been trying to find a tight upper bound for the series

$$S (x) = \sum_{k=1}^{\infty} \frac{(x_n-n)^k}{k!}$$

in terms of finite value $x\in \mathbb R$, where:

1- $\{x_n\}$ is a sequence of a real numbers.

2- $|x_n -(n+1)|\leq x$.

I think that the solution of the problem steps for expansion in Taylor series of

$$e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}$$

The difficulty of the problem lies in the fact that, while within the series there is the term $(x_n-n)^k$, the condition (2) is on $|x_n -(n+1)|$.

Answer 1

I would just write $$ |S(x)| = |e^{x_n-n}-1| =|e^{(x_n-(n+1))+1}-1| \leq e^{|(x_n-(n+1))+1|}-1\leq e^{|x_n-(n+1)|+1} -1\leq e^{|x|+1}-1 $$ Observe that $|x|=x$ in your question.

Answer 2

Hint

I suppose that you are very close since $$S (x) = \sum_{k=1}^{\infty} \frac{(x_n-n)^k}{k!}=\sum_{k=0}^{\infty} \frac{(x_n-n)^k}{k!}-1=e^{(x_n-n)}-1$$