We say a real valued sequence $(a_{n})_{n\geq0}$ is anti-Cauchy if $\forall\varepsilon>0$, there exists some $N\in\mathbb{N}$ such that : $$ |a_{n}-a_{m}|>\varepsilon\quad\forall m,n>N,\;m\neq n $$ For natural numbers K, we define the $K^{th}$ difference sequence $D_{i}(k)$ of a sequence $a_{n}$ and the $K^{th}$ sum sequence $S_{i}(k)$ of a sequence $a_{n}$ as follow : (note that $i\geq0$) $$ \begin{align} D_{i}(k)=& \begin{cases} D_{i}(0)=a_{i}\\ D_{i}(k+1)=D_{i+1}(k)-D_{i}(k)&\text{for $k\geq0$} \end{cases} \\ \\ \qquad S_{i}(k)=& \begin{cases} S_{i}(0)=a_{i}\\ S_{i}(k+1)=\displaystyle\sum_{r=0}^{i}S_{r}(k)&\text{for $k\geq0$} \end{cases} \end{align} $$ $1)$ Does there exist for each natural number $K$, a sequence such that its $k^{th}$ difference sequence is actually anti-Cauchy if and only if $k\leq K$ ?
$2)$ Does there exists a real valued sequence, not identically $0$ such that its $K^{th}$ sum sequence is never anti-Cauchy for any natural $K$?
UPDATE :
I would like to thank mathworker21 for his partial answer, I would also thank for T.H. Shehadi for his guess for part $(1)$ yet I am unsure if it is a correct answer or not I am still looking for a credible answer for $(1)$.
PROPOSITION :
T. H. Shehadi has offered an alternative definition for the $K^{th}$ difference sequence and $K^{th}$ summation sequence that would describe difference sum an in terms of partial sum I hope it could help in solving $(1)$ :
$(1)$ : Let Λ map sequences to sequences so that $\Lambda(a_{0})=a_{0}$ and $\Lambda(a_{n+1})=a_{n+1}−a_{n}$. We say that $\Lambda^{k}(a)$, i.e. $\Lambda$ composed with itself $k$ times, is the $k^{th}$ difference sequence of $a$.
$(2)$ : Let $\Omega$ map sequences to sequences such that $\Omega(a_{n})=a_{0}+a_{1}+\cdots +a_{n}$. We say that $\Omega^{k}(a)$ i.e. $\Omega$ composed with itself $k$ times, is the $k^{th}$ summation sequences of $a$ where $\Omega^{0}(a)=a$
The answer to (2) is yes.
Consider the pattern $x_1,x_2,\dots = 1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,1,2,3,4,5,6,\dots$.
Our real-valued sequence will start off with $a_1 = 1$. With $a_1,\dots,a_{n-1}$ chosen, we choose $a_n$ so that $S_n(x_k) = S_{n-1}(x_k)$, which is possible, since $S_{n-1}(x_k)$ just depends on $a_1,\dots,a_{n-1}$ and since $S_n(x_k)$ depends linearly on $a_n$.
We claim the $K^{\text{th}}$ sum of $(a_n)_n$ is anti-cauchy for all $K \ge 1$. Indeed, fix $K \ge 1$, and just note that, by construction, we have $S_n(K) = S_{n-1}(K)$ for infinitely many $n$.