# Show that given sequence is convergent.

Show that given sequence$\{S_n\}$ is convergent.

$$S_n = 1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+.....+\dfrac{1}{(n-1)!}$$

is convergent.

My input:

$S_n = 1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+.....+\dfrac{1}{(n-1)!}<1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{2^2}...=1+1+1=3$

$S_n<3$

Given sequence is increasing and we have proved that it's bounded above by $3$. Thus making it a convergent sequence. Is it a correct approach to solve this problem? Any other way I could've solved this?

• Seems correct and nice. – hamam_Abdallah Aug 23 '18 at 14:52
• @Cornman I didn't realise that we were taking the limit of $S_{n}$, my bad. – Bill Wallis Aug 23 '18 at 14:53
• Just to note some technicalities. On your inequality you have used $S_n$ to mean at least two different things - you need to call the right-hand side $T$ or something else. You shouldn't use $T_n$ (or $S_n$) for the infinite sum. You might want $S_n\lt T_n\lt T$ so that $\lim S_n \le T$. You have indicated a geometric series comparison, but you should give the general term to make this explicit. There are more delicate tests of convergence available, but you have taken advantage of the rapid convergence here to make an easy and quick comparison. – Mark Bennet Aug 23 '18 at 14:59
• @MarkBennet That was useful thanks. – Daman Aug 23 '18 at 15:01
• Yes this is right, but the explanation is a little short. One may wonder where $1+1+1$ comes from (not counting the typo). Would be good to say for $n>2$, $1/n!<2^{-n}$. – Yves Daoust Aug 23 '18 at 15:09

Use the ratio test, so: $$\frac{1}{(n+1)!} \div \frac{1}{n!}=\frac{n!}{(n+1)!}=\frac{1}{n+1}<1\forall n>0$$

• Ratio test for sequences is different from ratio test for series right? For sequences it is $\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=l;$ convergent if $l<|1|$ right ? – Daman Sep 2 '18 at 10:00

Yes, it is correct.

The standard way of proving that this sequence converges is to use the ratio test: if $(a_n)_{n\geqslant0}$ is a sequence of positive numbers such that $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}<1$, then the sequence $a_0+a_1+a_2+\cdots$ converges.

• The ratio test is essentially a comparison to a geometric series. – Yves Daoust Aug 23 '18 at 15:11
• @YvesDaoust Sure. – José Carlos Santos Aug 23 '18 at 15:14

Yes it is correct, as an alternative we have

$$\lim_{n\to \infty}S_n=\lim_{n\to \infty} \sum_{k=1}^n \frac 1{(k-1)!}=\sum_{k=1}^\infty \frac 1{(k-1)!}$$

which converges for example by comparison test with $\sum \frac 1{k^2}$.

Take $$u_n=1+\sum_{i=1}^n\frac{1}{i!}$$ and

$$v_n=u_n+\frac{1}{nn!}$$

It is easy to prove that $(u_n)$ and $(v_n)$ are adjacent and therefore convergent.

You can also use this to prove that the limit $e\notin \Bbb Q$.

An option:

Let $k\ge 2$.

$\dfrac{1}{k!} \le \dfrac{1}{k(k-1)} = \dfrac{1}{k-1} - \dfrac{1}{k}.$

We have

$\displaystyle {\sum_{k=2}^{n}} \dfrac{1}{k!} \le \displaystyle{ \sum_{k=2}^{n}}[ \dfrac{1}{k-1}-\dfrac{1}{k}] =$

$1- \dfrac{1}{n}.$

1) $S_n=\displaystyle{ \sum_{k=0}^{n-1}}\dfrac{1}{k!}$ is strictly increasing,

2) $S_n$ is bounded above by $3$ (why?),

hence convergent.