Prove that if $m>n$, then $|u_m - u_n| \leq |v_{m,1/2} - v_{n,1/2}|$ Given $u_n = \sum_{k=0}^n \frac{1}{k!}$ and $v_{n, 1/2} = \sum_{k=0}^n \frac{1}{2^k}$.
Prove that $|u_m - u_n| \leq |v_{m,1/2} - v_{n,1/2}|$ for any $m > n$. I tried this but I don't see how to prove it. Here is what I did:
$|u_m - u_n| = |\sum_{k=0}^m \frac{1}{k!} - \sum_{k=0}^n\frac{1}{k!}| = |\sum_{k=n+1}^m \frac{1}{k!}|$
Note that: $\frac{1}{k!} = \frac{1}{2}\frac{1}{3}\dots\frac{1}{k} \leq \frac{1}{2}\frac{1}{2}\dots\frac{1}{2} = \frac{1}{2^{k-1}}$, which implies that $\sum_{k=n+1}^m \frac{1}{k!} \leq \sum_{k=n+1}^m \frac{1}{2^{k-1}} = 2\sum_{k=n+1}^m\frac{1}{2^{k}}$.
So we get $|u_m - u_n| = |\sum_{k=n+1}^m \frac{1}{k!}| \leq |2\sum_{k=n+1}^m\frac{1}{2^{k}}| = |2||\sum_{k=n+1}^m\frac{1}{2^{k}}| = 2|v_{m,1/2} - v_{n,1/2}|$.
Therefore $|u_m - u_n| \leq 2|v_{m,1/2} - v_{n,1/2}|$ - which is NOT what I'm trying to prove. Clearly, I'm doing something wrong here. Any help? I've also proved $v_{m, r} - v_{n,r}\leq \frac{r^{q+1}}{1-r}$, but I don't see how that helps.
 A: Thanks to xhsbm for pointing out that when $a \in {1,2,3}$ then $2^a > a!.$  The proof below is valid for $n \geq 4$, under the assumption that $m > n$.  I will have to edit my answer, treating $n < 4$ as a special case.

Assuming that $m > n$, the cases of $n=1, n=2, n=3$ must be
empirically checked.
It's easy to see that if $n=1$, the conjecture is always false, because the 1st term on the LHS is $\frac{1}{2!}.$
For $n=2$, I found that empirically that the conjecture is false for $m=3$ or $m=4$, and true for $m \geq 5$.
For $n=3$, the conjecture is always true, since the first term on the LHS is $\frac{1}{4!}$ and the first term on the RHS is $\frac{1}{2^4}.$

I am assuming that $m,n$ are positive integers.
Your work was good, but you accidentally went down the wrong path.
In general, since $2^a < a!,$
$[E_1]: ~\frac{1}{a!} < \frac{1}{2^a}.$
WLOG $m>n.$
You are being asked to prove that
$\frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \frac{1}{(n+3)!} + \cdots
+ \frac{1}{(m)!}.$
is less than or equal to
$\frac{1}{2^{(n+1)}} + \frac{1}{2^{(n+2)}} + \frac{1}{2^{(n+3)}} + \cdots + \frac{1}{2^{(m)}}.$
You can use $[E_1]$ above to compare the two sums term by term.
