Showing that $u_n=\sum_{k=0}^{n}\frac{1}{k!}$ is Cauchy Consider $u_n=\sum_{k=0}^{n}\frac{1}{k!}$. I would like to show this sequence is Cauchy, that is
$\forall\epsilon>0, \exists N\in\mathbb{N} : n,m\geq N\implies \lvert u_n - u_m\rvert\leq\epsilon$
My attempt : Consider $\epsilon>0$ and suppose $n<m$. For what $N$ can we get $\lvert u_n - u_m\rvert\leq\epsilon\Longleftrightarrow\lvert\sum_{k=n+1}^{m}\frac{1}{k!}\rvert\leq\epsilon$ ?
This attempt is totally wrong : see the answer kindly suggested by Michael using the comparison with a Cauchy sequence
Clearly if we have $\forall k\in(n+1, m) : \lvert\frac{1}{k!}\rvert = \frac{1}{k!}\leq\frac{\epsilon}{m-n}$ we are done because it would lead to $\lvert\sum_{k=n+1}^{m}\frac{1}{k!}\rvert\leq\sum_{k=n+1}^{m}\lvert\frac{1}{k!}\rvert\leq\epsilon$.
This means that we need to have $k!\geq\frac{m-n}{\epsilon}$.
So we take $N$ to be the ceiling function of $\frac{m-n}{\epsilon}$.
Is this seems correct to you ?
Thank you a lot
 A: So, let's think back to the definition of a Cauchy sequence:
A sequence $\{u_n\}$ of real numbers is Cauchy if for every $\epsilon > 0$, there exists $N>0$ such that if $n,m>N$, then $|u_n - u_m| < \epsilon$
Note that in the definition, the value of $N$ does not depend on the values of $n$ or $m$. It should only depend on the value of $\epsilon$ that you chose. In other words, we first have to find the value of $N$ (based ONLY on the value of $\epsilon$ that was chosen), and only once we've done so can we start choosing values of $n$ or $m$ that are larger than $N$.
In your argument, by letting $N = \lceil\frac{m-n}{\epsilon} \rceil$, you've essentially done the opposite. Your value of $N$ depends on the values of $n$ and $m$, which is a problem.
Instead, let's try a different approach. You were given a hint in the comments that $\frac{1}{n!} \leq 2^{-n}$. Let's try to use this fact. So, we pick $\epsilon > 0$. We do want to find a value of $N$ so that $\lvert \sum_{k = n+1}^m \frac{1}{k!} \rvert < \epsilon$. Note that all of the terms in the sum are non-negative, so we can even eliminate the absolute values to make things cleaner.
Observe, by the hint, that $$\sum_{k = n+1}^m \frac{1}{k!} \leq \sum_{k = n+1}^m 2^{-k}$$
This gives us a hint that perhaps the geometric series $\sum_{k = 0}^\infty 2^{-k}$ will be involved. We know that the geometric series converges, so its sequence of partial sums $T_m := \sum_{k = 0}^m 2^{-k}$ must converge. Since this is a sequence of real numbers, this is the same as saying that the sequence $\{T_m\}$ of partial sums is Cauchy! Thus, for the $\epsilon$ we chose at the very beginning, there exists a value of $N$, only depending on $\epsilon$, such that if $n,m > N$, then $\lvert T_m - T_n \rvert < \epsilon$. So for $m > n > N$, we will have $$|u_n - u_m | = \sum_{k = n+1}^m \frac{1}{k!} \leq \sum_{k = n+1}^m 2^{-k} = |T_m - T_n| < \epsilon$$
which is the desired result.
