# Show that $\{S_n=\sum_{k=1}^n \frac{1}{k!}\}$ is convergent by showing that $S_n$ is a Cauchy sequence.

Show that $$\{S_n=\sum_{k=1}^n \frac{1}{k!}\}$$ is convergent by showing that $S_n$ is a Cauchy sequence.

The hint given is $r!\geq2^{r-1}$ and $\sum_{r=1}^{\infty} \frac{1}{2^r}=2$

I know the definition of Cauchy sequence that is for each $\epsilon>0$ there exists a natural number such that $m,n\geq N$ implies that $|s_n-s_m|<\epsilon$

Here I also want to ask that what is defined by the $s_m$?

• $s_m$ is the $m$-th partial sum of the sequence $\{a_n\}$ (say). $s_m=a_1+a_2+\cdots+a_m$. – Sayan Nov 28 '14 at 14:31

For any $n\geq 4$ we have $n!>2^n$, hence for any $n\geq 4$ and for any $m\geq n$ we have: $$0\leq S_m-S_n \leq \frac{1}{2^n}+\ldots+\frac{1}{2^m} \leq \sum_{k\geq n}\frac{1}{2^k}=\frac{1}{2^{n-1}}.$$ The last line trivially gives that $\{S_n\}_{n\in\mathbb{N}}$ is a Cauchy sequence.

• The last line guarantees the Cauchy sequence because $1/(2^{n-1})$ must be less than $\epsilon$? – Alan Wang Nov 28 '14 at 14:43
• @AlanWang: exactly. As soon as $n>1-\log_2\varepsilon$, we have $\frac{1}{2^{n-1}}<\varepsilon$. – Jack D'Aurizio Nov 28 '14 at 14:59

Hint:

Without loss of generality, let $n > m$. Then

$$|s_n - s_m| = \left| \sum_{k=1}^n \frac{1}{k!} - \sum_{k=1}^m \frac{1}{k!} \right| = \left| \sum_{k=m+1}^n \frac{1}{k!} \right| = \sum_{k=m+1}^n \frac{1}{k!} \leq \sum_{k=m+1}^n \frac{1}{2^{k-1}}$$

Now can you show this last sum can be made arbitrarily small?