Use the Cauchy convergence criterion to show that the following sequence convergences (without quoting results for the series representation of the transcendental number $ e $).
$$ \sum_{k=1}^n \frac 1{k!} $$
Hint: $ k! \geq k(k-1) $ for $ k \geq 2$
How do I go about solving this problem?
This is my working so far:
Let $m>n$. (not sure about this)
$|\sum_{k=1}^m \frac 1{k!} - \sum_{j=1}^n \frac 1{j!}| = |\frac 1{1!}+\frac 1{2!}+...+\frac 1{m!}-\frac 1{1!}-\frac 1{2!}-...-\frac 1{n!}|=|\frac 1{(n+1)!}+\frac 1{(n+2)!}+...+\frac 1{(m-1)!}+\frac 1{m!}| < |\frac 1{n(n+1)}+\frac 1{(n+1)(n+2)}+...+\frac 1{(m-1)(m-2)}+\frac 1{m(m-1)}|$
I don't know what I should do next though. If someone can help me out I'd really appreciate it!
Next step: telescopic series. From where you left:
$$\sum_{k=0}^p\frac1{(n+k)(n+k+1)}=\frac1n-\frac1{n+1}+\frac1{n+1}-\frac1{n+2}+\ldots+\frac1{n+p}-\frac1{n+p+1}=$$
$$=\frac1n-\frac1{n+p+1}\xrightarrow[n\to\infty]{}0$$
and thus for any $\;\epsilon>0\;$ you can choose $\;p>n\;$ such that the above sum is less than $\;\epsilon\;$, and this is Cauchy.