Consider the series: $$\sum_{k=0}^{\infty}\frac{1}{k!}$$
Prove that the sequence of partial sums ($s_{n}=\sum_{k=0}^{n}\frac{1} {k!}$) $n>0$ is monotonically increasing.
My approach:
$$s_{1}= 1,\quad s_{2}= 1 + \frac{1}{2},\quad s_{3}= 1 + \frac{1}{2}+ \frac{1}{6}$$
$$\implies s_{1}<s_{2}<s_{3}$$
$$\implies \frac{1}{k!}< \frac{1}{k!(k+1)}< \frac{1}{k!(k+1)(k+2)}$$
Then by definition of monotonically increasing sequence the above mentioned property holds.
https://en.wikipedia.org/wiki/Sequence#Increasing_and_decreasing
Now I'm just not sure if that's enough to prove this.
Do I have to show by induction that $\frac{1}{k!}<\frac{1}{k!(k+1)}$ holds or is it clear?
Best Answer
You could prove this by induction.
First, prove the base case: the summation when $n=2$ is greater than the summation when $n=1$
Then, assume this is true for all $m<n$.
Need to show this holds for $n+1$ and $n$.
$\sum_{k=0}^{n+1}\frac{1}{k!}=(\sum_{k=0}^{n}\frac{1}{k!})+\frac{1}{(1+n)!}>\sum_{k=0}^{n}\frac{1}{k!}$
So it holds for $n+1$ and $n$.
Thus by induction, your proposition holds.