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.


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?

  • 2
    $\begingroup$ How about just each summand is positive? $\endgroup$ – vadim123 May 31 '16 at 18:02
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    $\begingroup$ It's enough to show that $s_{k+1}-s_k>0$ is true for all $k\in\mathbb Z^+$, which is equivalent to $\frac{1}{(k+1)!}>0$ for all $k\in\mathbb Z^+$, which is true. $\endgroup$ – user236182 May 31 '16 at 18:03
  • $\begingroup$ I think this is trivially true for any series that is the sum of non-negative elements... $\endgroup$ – DonAntonio May 31 '16 at 18:48

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 $n$.


So it holds for $n+1$ and $n$.

Thus by induction, your proposition holds.


The key aspect is to generalize this for all $n \in \mathbb{N} \cup \{0\}.$ If we wanted to prove that for some arbitrary $n \in \mathbb{N} \cup \{0\},$

$$s_{n+1} > s_n$$ $$\leftarrow \sum_{k=0}^{n+1} \frac{1}{k!} > \sum_{k=0}^{n} \frac{1}{k!}$$ $$\leftarrow \frac{1}{(n+1)!} + \sum_{k=0}^n \frac{1}{k!} > \sum_{k=0}^n \frac{1}{k!}$$ $$\leftarrow \frac{1}{(n+1)!} > 0.$$

We will call our last statement our condition. Given that $1 > 0,$ and $(n+1)!$ is positive since $n \in \mathbb{N} \cup \{0\},$ Then $\frac{1}{(n+1)!}$ must be positive and so $\frac{1}{(n+1)!} > 0.$ Hence, our condition holds.


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