We are asked to evaluate the summation of the expression $n-1$ from $n=c_1$ to $\xi$. It is likely that $c_1$ represents a constant and $\xi$ refers to infinity. Thus, we want to evaluate $\sum_{n=1}^{\infty} (n-1)$.

1. Problem Description

We are asked to evaluate the summation of the expression $n-1$ from $n=c_1$ to $\xi$. It is likely that $c_1$ represents a constant and $\xi$ refers to infinity. Thus, we want to evaluate $\sum_{n=1}^{\infty} (n-1)$.

2. Solution Steps

The summation can be rewritten as:

$\sum_{n=1}^{\infty} (n-1) = \sum_{n=1}^{\infty} n - \sum_{n=1}^{\infty} 1$

The sum $\sum_{n=1}^{\infty} n$ is the sum of all positive integers, which diverges to infinity.

The sum $\sum_{n=1}^{\infty} 1$ is also a divergent series.

We can write out the first few terms of the series:

When $n=1$, $n-1 = 1-1 = 0$.

When $n=2$, $n-1 = 2-1 = 1$.

When $n=3$, $n-1 = 3-1 = 2$.

When $n=4$, $n-1 = 4-1 = 3$.

So, the series is $0 + 1 + 2 + 3 + ...$, which is equivalent to the sum of all non-negative integers.

Since the terms of the sequence $(n-1)$ do not converge to 0 as $n$ approaches infinity, the series $\sum_{n=1}^{\infty} (n-1)$ diverges.

Specifically, $\sum_{n=1}^{N} (n-1) = \sum_{n=0}^{N-1} n = \frac{(N-1)(N-1+1)}{2} = \frac{(N-1)N}{2}$.

As $N$ goes to infinity, this expression also goes to infinity.

3. Final Answer

The series diverges to infinity.