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.

  • $\begingroup$ Could you write it as the point where $1 - \int_1^x \frac{1}{t} dt = 0$, and then use Newton's method or something? $\endgroup$ – andybenji Dec 9 '12 at 7:12
  • $\begingroup$ @andybenji: Thanks for your idea, but I'm still not sure how to apply that. $\endgroup$ – Chan Dec 9 '12 at 7:21
  • $\begingroup$ Did the question say to use that method, or did you think that was the right direction to go? $\endgroup$ – andybenji Dec 9 '12 at 7:23
  • $\begingroup$ The question did not state a specific method, so I guess I need to use the definition of $\log$ and $\exp$ since I've just learned it. $\endgroup$ – Chan Dec 9 '12 at 7:28

$e=\lim_{n\to \infty}(1+\frac1n)^n$

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 $1+\frac1{1!}+\frac1{2!}+\frac1{3!}+\cdots>1+1+0.5=2.5$





$e=1+\frac1{1!}+\frac1{2!}+\frac1{3!}+\cdots$ $<1+1+\frac12+\frac1{2^2}+\frac1{2^3}+\cdots$ $=1+(1+\frac12+\frac1{2^2}+\frac1{2^3}+\cdots)=1+\frac{1}{1-\frac12}=3$ as the terms inside parenthesis forms an infinite geometric series with the common ratio $=\frac12,$ the 1st term being $1$


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+\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+\frac{1}{2}+\frac{1}{6}+\frac{1}{8}\Rightarrow e\le 1+\frac{1}{2}+\frac{1}{6}+\frac{1}{8}<3 \end{equation}

  • $\begingroup$ Can't we only conclude from $(1+1/n)^n < 3$ that $e \leq 3$? $\endgroup$ – Antonio Vargas Dec 9 '12 at 7:22
  • 1
    $\begingroup$ @AntonioVargas Now fixed $\endgroup$ – Nameless Dec 9 '12 at 7:52
  • $\begingroup$ Nicely done :-) $\endgroup$ – Antonio Vargas Dec 9 '12 at 7:57

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

$$\frac{5}{2} < \left(1 + \frac{1}{6}\right)^6 \leq e \leq \left(1-\frac{1}{6}\right)^{-6} < 3.$$

  • $\begingroup$ Nice proof straight from the definition of $\log$ and $\exp$. Thanks a lot. $\endgroup$ – Chan Dec 9 '12 at 9:38

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.

  • $\begingroup$ I did only the other side, but it turns out that for that you want to choose $m$ so that $H_{3m}-H_m>1$, where $H_n$ is the $n$-th harmonic number, and then form the righthand sum over the partition of $[1,3]$ into $2m$ intervals of length $\frac1m$; $m=4$ works. $\endgroup$ – Brian M. Scott Dec 9 '12 at 9:29
  • $\begingroup$ +1. Very straightforward and a fairly small number of intervals suffice. $\endgroup$ – WimC Dec 9 '12 at 9:49

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.)


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, $$e^{2/3} \leq 2 \implies e \leq 2^{3/2} = 2 \sqrt{2}$$


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 $$ \begin{align} \frac52=1+1+\frac12< e &=1+\frac11+\frac1{1\cdot2}+\frac1{1\cdot2\cdot3}+\frac1{1\cdot2\cdot3\cdot4}+\dots\\ &<1+\frac11+\frac1{1\cdot2}+\frac1{1\cdot2\cdot2}+\frac1{1\cdot2\cdot2\cdot2}+\dots\\ &=3\tag{6} \end{align} $$


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