Prove that $5/2 < e < 3$? 
Prove that $$\dfrac{5}{2} < e < 3$$

By the definition of $\log$ and $\exp$, 
$$1 = \log(e) = \int_1^e \dfrac{1}{t} dt$$
where $e = \exp(1)$.     
So given that $e$ is unknown, how could I prove this problem? I think I'm missing some important facts that could somehow help me rewrite $e$ in some form of $3$ and $5/2$. Any idea would be greatly appreciated.
 A: As we know $e=\lim_{n\to +\infty}\left(1+\dfrac{1}{n}\right)^n$. That sequence is increasing (why?) and so $$\frac{5}{2}<\left(1+\dfrac{1}{10}\right)^{10}<e$$
In addition, for  $n=1,2...$, by the Binomial Theorem,
\begin{gather}\left(1+\dfrac{1}{n}\right)^n=\sum_{k=0}^{n}\binom{n}{k}\left(\dfrac{1}{n}\right)^k=\sum_{k=0}^{n}\frac{1}{k!}\frac{n!}{(n-k)!}\frac{1}{n^k}=
1+\sum_{k=1}^{n}\frac{1}{k!}\frac{n(n-1)...(n-k+1)}{n^k}= \notag\\
1+\sum_{k=1}^{n}\frac{1}{k!}\frac{n}{n}\frac{n-1}{n}...\frac{n-k+1}{n}=
1+\sum_{k=1}^{n}\frac{1}{k!}\left(1-\frac{1}{n}\right)...\left(1-\frac{k-1}{n}\right)<1+\sum_{k=1}^{n}\frac{1}{k!}\end{gather}
(you can skip the above if you know $e=\sum_{n=0}^{\infty}\frac{1}{n!}$)
Since $k\ge 4\Rightarrow k!\ge 2^{k}$,
\begin{equation}\left(1+\dfrac{1}{n}\right)^n<1+1+\frac{1}{2}+\frac{1}{6}+\sum_{k=4}^{n}\frac{1}{2^{k}}\end{equation}
For the last sum, observe that
\begin{equation}\sum_{k=4}^{n}\frac{1}{2^{k}}=\frac{1}{16}\frac{\frac{1}{2^{n-4}}-1}{\frac12-1}=\frac{1}{8}-\frac{1}{2^{n-1}}<\frac{1}{8}
\end{equation}
and thus,
\begin{equation}\left(1+\dfrac{1}{n}\right)^n<1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{8}\Rightarrow e\le 1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{8}<3
\end{equation}
A: From $$\log(x) = \int_1^x\frac{dt}{t}$$ and using a constant lower and upper bound for $1/t$ on the interval $[1, x]$ it follows that $$1 - \frac{1}{x} \leq \log(x) \leq x - 1$$ for all $x > 0$.  Taking inverse functions this becomes $$1 +x \leq e^x \leq \frac{1}{1-x}$$ for all $x < 1$ (the lower bound holds for all $x \in \mathbb{R}$, you might want to draw a picture).  Take $n \geq 1$, substitute $x \leftarrow x/n$ and raise to the power $n$ to get $$\left(1 + \frac{x}{n}\right)^n \leq e^{\frac{x}{n}n} = e^x \leq \left(1 - \frac{x}{n}\right)^{-n}$$ for all $x < n$.  For $x=1$ and $n=6$ this becomes
$$\frac{5}{2} < \left(1 + \frac{1}{6}\right)^6 \leq e \leq \left(1-\frac{1}{6}\right)^{-6} < 3.$$
A: Use a Riemann sum to guarantee that 
$$
\int_1^{\frac{5}{2}} \frac{dt}{t} < 1.
$$
(use a left hand sum to guarantee that the Riemann sum overestimates the integral). Similarly with a right hand sum for the other inequality.
A: Method 1:
Using the power series for $\log(1+x)$, we get
$$
\begin{align}
\log\left(1+\frac1n\right)
&=-\log\left(1-\frac1{n+1}\right)\\
&=\frac1{n+1}+\frac1{2(n+1)^2}+\frac1{3(n+1)^3}+\dots\tag{1}
\end{align}
$$
Multiply $(1)$ by $n+1$:
$$
(n+1)\log\left(1+\frac1n\right)=1+\frac1{2(n+1)}+\frac1{3(n+1)^2}+\dots\tag{2}
$$
$(2)$ is obviously decreasing in $n$, therefore $\displaystyle\left(1+\frac1n\right)^{\large n+1}$ is decreasing in $n$.
Multiply $(1)$ by $n$:
$$
\begin{align}
n\log\left(1+\frac1n\right)
&=((n+1)-1)\log\left(1+\frac1n\right)\\
&=1-\frac1{1\cdot2(n+1)}-\frac1{2\cdot3(n+1)^2}-\dots\tag{3}
\end{align}
$$
$(3)$ is obviously increasing in $n$, therefore $\displaystyle\left(1+\frac1n\right)^{\large n}$ is increasing in $n$.
Therefore, since $\displaystyle e=\lim_{n\to\infty}\left(1+\frac1n\right)^{\large n}$, we have
$$
\left(1+\frac1n\right)^{\large n}< e<\left(1+\frac1n\right)^{\large n+1}\tag{4}
$$
Let $n=6$, then
$$
\frac52<\left(\frac76\right)^{\large 6}< e<\left(\frac76\right)^{\large 7}<3\tag{5}
$$
Method 2:
Since $\displaystyle e=\sum_{n=0}^\infty\frac1{n!}\ $, we have
$$
\begin{align}
\frac52=1+1+\frac12< e
&=1+\frac11+\frac1{1\cdot2}+\frac1{1\cdot2\cdot3}+\frac1{1\cdot2\cdot3\cdot4}+\dots\\
&<1+\frac11+\frac1{1\cdot2}+\frac1{1\cdot2\cdot2}+\frac1{1\cdot2\cdot2\cdot2}+\dots\\
&=3\tag{6}
\end{align}
$$
A: $e=\lim_{n\to \infty}(1+\frac1n)^n$
the rth term $t_r=\frac{n(n-1)(n-2)\cdots(n-r+1)}{r!n^r}=\frac1{r!}\prod_{0\le s<r}(1-\frac sn)$ wheer $1\le r<\infty$
So, $\lim_{n\to \infty}t_r=\frac1{r!}$
So, $e=1+\frac1{1!}+\frac1{2!}+\frac1{3!}+\cdots$
But $1+\frac1{1!}+\frac1{2!}+\frac1{3!}+\cdots>1+1+0.5=2.5$
Again, 
$3!=1.2.3>1.2.2=2^2$
$4!=1.2.3.4>1.2.2.2=2^3$
So, 
$e=1+\frac1{1!}+\frac1{2!}+\frac1{3!}+\cdots$
$<1+1+\frac12+\frac1{2^2}+\frac1{2^3}+\cdots$
$=1+(1+\frac12+\frac1{2^2}+\frac1{2^3}+\cdots)=1+\frac{1}{1-\frac12}=3$
 as the terms inside parenthesis forms an infinite geometric series with the common ratio $=\frac12,$ the 1st term being $1$
A: We know that $$e = \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n$$
This is a Cauchy sequence (why?), so for any $\varepsilon > 0$ there exists $N = N(\varepsilon) \in \mathbb{N}$ satisfying $|e_n - e_m| < \varepsilon$.  (Here $e_n$ is the $n$th term of the sequence).  Then, it must converge, and we may take $\varepsilon$ to be $\frac{1}{100}$, or something like that.  Find $N(\frac{1}{100})$, and then you know that all the values lie in a ball of radius $\frac{1}{100}$ about $e_N$, and thus the limit lies in the closure of this ball.  (The key fact is that we may shrink this ball so that $5/2$ and $3$ are far away from it.)
A: We can prove a slightly tighter upper-bound as follows.
Let $$H_n = 1 + \frac1{2} + \cdots + \frac1{n}$$
$$H_{2n} - H_{n} = \frac1{n+1} + \frac1{n+2} + \cdots + \frac1{2n}$$
By AM-HM, we have $$\frac{H_{2n} - H_n}{n} \geq \frac{n}{(n+1) +(n+2) + (n+3) + \cdots + 2n} = \frac{n}{n^2 + \frac{n(n+1)}{2}}$$
$$H_{2n} - H_n \geq \frac{2n}{3n+1}$$
$$(H_{2n} - \log(2n)) - (H_n - \log(n)) + \log(2) \geq \frac{2n}{3n+1}$$
Take limit $n \rightarrow \infty$ and note that $\displaystyle \lim_{n \rightarrow \infty} (H_n - \log(n)) = \gamma < \infty$ to get $\log(2) \geq \frac{2}{3}$.
Hence, $$e^{2/3} \leq 2 \implies e \leq 2^{3/2} = 2 \sqrt{2}$$
A: The goal here is simply to show
$$\int_1^{5/2}{dt\over t}\lt1\lt\int_1^3{dt\over t}$$
using properties of the definite integral and the function $f(t)=1/t$, which, for $t\ge1$, is decreasing and concave up. Concavity tells us the Trapezoid Rule gives an upper bound on the area beneath the curve, hence (using a subsitution $u=2t$ to get things started)
$$\begin{align}
\int_1^{5/2}{dt\over t}=\int_2^5{du\over u}&\lt{1\over2}(f(2)+2f(3)+2f(4)+f(5))\\
&={1\over2}\left({1\over2}+{2\over3}+{2\over4}+{1\over5} \right)\\
&\lt{1\over2}\left({1\over2}+{4\over5}+{1\over2}+{1\over5}\right)\\
&=1
\end{align}$$
while the decreasing nature of $1/t$ tells us a Riemann sum using right hand endpoints gives a lower bound, hence, splitting $[1,3]$ into eight segments of width $1/4$, we have
$$\begin{align}
\int_1^3{dt\over t}&\gt{1\over4}(f(5/4)+f(6/4)+f(7/4)+f(8/4)+f(9/4)+f(10/4)+f(11/4)+f(12/4))\\
&={1\over5}+{1\over6}+{1\over7}+{1\over8}+{1\over9}+{1\over10}+{1\over11}+{1\over12}\\
&=\left({1\over5}+{1\over10}\right)+\left({1\over6}+{1\over12}\right)+\left({1\over7}+{1\over8}\right)+\left({1\over9}+{1\over11}\right)\\
&={3\over10}+{1\over4}+{15\over56}+{20\over99}\\
&\gt{3\over10}+{1\over4}+{14\over56}+{20\over100}\\
&={3\over10}+{1\over4}+{1\over4}+{2\over10}\\
&=1
\end{align}$$
(It can help to draw a picture of the curve, trapezoids, and rectangles to see the comparisons.)
Remark: Strictly speaking, there is no need for the extra inequalities that replace certain fractions with others -- you can, if you like, just do the calculations. But whenever possible, I like to write things up in such a way that I can reread things doing all the arithmetic in my head.
A: Let $$e_n=\left(1+\frac{1}{n}\right)^n,\quad s_n=\sum_{k=0}^{n}\frac{1}{k!}\ (0!=1),$$
then $e_1=2=s_1,e_2=\frac{9}{4}<\frac{5}{2}=s_2$, when $n>2$,
\begin{align*}
\left(1+\frac{1}{n}\right)^n
 &=1+1+\sum_{k=2}^{n}\frac{1}{k!}\left(1-\frac{1}{n}\right)\cdots\left(1-\frac{k-1}{n}\right)\\
 &<1+1+\sum_{k=2}^{n}\frac{1}{k!}\\
 &=1+1+\frac{1}{2}+\sum_{k=3}^{n}\frac{1}{k!}\\
 &\leq\frac{5}{2}+\sum_{k=3}^{n}\frac{1}{k(k-1)(k-2)}\\
 &=\frac{5}{2}+\frac{1}{2}\sum_{k=3}^{n}\frac{1}{k-1}\left(\frac1{k-2}-\frac1k\right)\\
 &=\frac{5}{2}+\frac{1}{2}\sum_{k=3}^{n}\frac{1}{(k-1)(k-2)}-\frac{1}{2}\sum_{k=3}^{n}\frac{1}{k(k-1)}\\
 &=2.75+\frac{1}{2}\left(\frac1n-\frac{1}{n-1}\right)\\
 &<2.75.
\end{align*}
Let $n\to \infty$, we have $$e\leq 2.75<3.$$
