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Suppose $(a_n)$ is a sequence such that $$a_n=\frac{1!+2!+\cdots+n!}{n!} \, .$$ Show that $\lim_{n \rightarrow \infty}{a_n}=1$. My attempt is to formulate an inequality and then use the Squeeze Theorem. Since we know that $$\frac{1}{n!}+\frac{2!}{n!}+\cdots+1>1$$ as $\frac{i!}{n!}>0$ for all $1 \leq i \leq n-1$, we have $a_n \geq1$. Then I'm stuck at formulating another side. Can anyone help?

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The Stolz–Cesàro theorem works well. – Jonas Meyer Feb 3 '13 at 4:42
An intuitive way: Suppose it converges, then $a_{n+1}=a_n/(n+1)+1$. Taking limits then yields the result. – awllower Feb 3 '13 at 4:54
@awllower: As Brian had pointed out, using that equation, it suffices to show that $(a_n)$ is bounded. – Jonas Meyer Feb 3 '13 at 5:14
@JonasMeyer Indeed. Thanks for pointing out this. – awllower Feb 3 '13 at 6:43
up vote 17 down vote accepted

Each of $1!,2!,...,(n-2)!$ is at most $(n-2)!$. Since there are $n-2$ such terms you have $$\frac{1! + 2! + \cdots + n!}{n!} \leq \frac{(n-2)(n-2)! + (n-1)! + n!}{n!}$$ $$= \frac{n-2}{n(n-1)} + \frac{1}{n} + 1$$ The limit of this as $n \rightarrow \infty$ is $1$, which is what you need.

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We will prove that: $$1!+2!+\cdots+n!\le 2\cdot n!$$ If $n=1$, it is trivial. And if $1!+2!+\cdots+n!\le 2\cdot n!$ holds, then $$1!+2!+\cdots+n!+(n+1)! \le 2\cdot n! +(n+1)! =(n+3)n! \le 2(n+1) n!$$ for all $n\ge 1$.

So we get $$\frac{1!+2+\cdots+(n-1)!+n!}{n!}\le \frac{2(n-1)!+n!}{n!}=1+\frac{2}{n}.$$

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what motivates you to think of proving $1!+2!+...+n! \leq 2.n!$ ? – Idonknow Feb 3 '13 at 4:43
@Idonknow Because $(n-1)!/n!=1/n$ and factorial is increasing very fast. – Hanul Jeon Feb 3 '13 at 4:58
Very elegant this answer. Moreover, that inequality is good to be kept in mind (+1). Actually, it may be rewritten as $1!+2!+\cdots+(n-1)!\le n!$ – user 1618033 Feb 3 '13 at 8:58

As Jonas Meyer suggested, the Stolz–Cesàro theorem might be another option. $$\lim_{n\to\infty}\frac{1!+2!+\cdots+n!}{n!}=\lim_{n\to\infty}\frac{(n+1)!}{(n+1)!-n!}=\lim_{n\to\infty}\frac{n+1}{n}=1.$$


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HINT Prove that $$\dfrac{k!}{n!} < \dfrac{C}{n^2}$$ for $k \in \{1,2,3,\ldots,n-2\}$. Move the cursor over the gray area for a complete answer.

Note that for $n \geq 2$ and $k \leq n-2$, we have $\dfrac{k!}{n!} \leq \dfrac1{n(n-1)} < \dfrac2{n^2}$. Hence,$$1 < \sum_{k=1}^{n} \dfrac{k!}{n!} < \dfrac2n + \dfrac{(n-1)!}{n!} + \dfrac{n!}{n!} = \dfrac3n + 1$$Hence, the limit is $1$.

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You have

$$\begin{align*} a_n&=\sum_{k=0}^{n-1}\frac{(n-k)!}{n!}\\ &=1+\frac1n\sum_{k=1}^{n-1}\frac{(n-k)!}{(n-1)!}\\ &=1+\frac1n\sum_{k=0}^{n-2}\frac{(n-1-k)!}{(n-1)!}\\ &=1+\frac1na_{n-1}\;. \end{align*}$$

If you can find an $M$ such that $a_n\le M$ for all $n$, then you’ll have $a_n\le 1+\frac{M}n$ for all $n$, which squeezes down nicely.


$$a_n=1+\underbrace{\frac1n+\frac1{n(n-1)}=\ldots+\frac1{n!}}_{n-1\text{ terms}}\;;$$

can you put a constant upper bound on that?

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