If $a_n=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}$, show that $2\le \lim_{n\to \infty} a_n\le 3$.

Question

If $ a_n=\cfrac{1}{0!}+\cfrac{1}{1!}+\cfrac{1}{2!}+\dots+\cfrac{1}{n!}$, show that $2\le \lim_{n\to \infty} a_n\le 3$.

Answer 1

Hint:

\begin{align} \sum_{k=0}^n \frac{1}{k!}&=2+\sum_{k=2}^n \frac{1}{k!}\\ &\leq 2+\sum_{k=2}^n\frac{1}{k(k-1)}\\ &=\ldots\\ &=2+\sum_{k=2}^n \frac{1}{k-1}-\sum_{k=2}^n\frac{1}{k}\\ &=2+1 - \text{?}\\ \end{align}

I deliberately left gaps for you to fill in.

Answer 2

First let's change the question a little:

Set $b_n=a_{n+4}$ so we get $2\le \lim_{n\to\infty}a_n\le 3$ to be $2\le \overbrace{a_0+a_1}^2+a_2+a_3+\lim_{n\to\infty}b_n\le 3\iff 0\le a_2+a_3+\lim_{n\to\infty}b_n\le 1$

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Now we notice that for all $n>3$ we have $\frac1{2^n}>\frac1{n!}$, so we have$$0\le a_2+a_3+\lim_{n\to\infty}b_n< a_2+a_3+\lim_{n\to\infty}\left(\frac1{2^4}+\frac1{2^4}+\cdots+\frac1{2^n}\right)$$ I'll leave the proof that the last part(the geometric series) is equal to $\frac18$. From this we get$$0\le a_2+a_3+\lim_{n\to\infty}b_n< a_2+a_3+\frac18=\frac{19}{24}<1$$Hence $2\le \lim_{n\to\infty}a_n< 3$ we can change the $<$ to $\le$ without loosing anything to get $2\le \lim_{n\to\infty}a_n\le 3$

Answer 3

An option:

$2 \lt a_n =$

$1+1+1/2!+1/3! + 1/n! \lt $

$(2) +(1/2 +1/2^2+...+1/2^n) \lt$

$(1+1) + (1/2+1/2^2+1/2^3.....)$.

The sum in the second parenthesis above is the infinite geometric series, first term $1/2$, constant ratio $1/2.$

What is it's sum?