How to prove that $e^k > k$?

Question

I'm working on series and I'm always stuck at one point where I dont know how to prove that $$e^k > k$$

I'm trying to show by a comparaison test that

$$\sum_{n=1}^\infty \frac{n}{e^n}$$ converges, so I try to compare it to the geometric sum

$$\sum_{n=1}^\infty \frac{1}{e^{n/2}}$$ which converges because it is equivalent to $$\sum_{n=1}^\infty \left(\frac{1}{e^{1/2}}\right)^n= \frac{1}{e^{1/2}}\sum_{n=1}^\infty \left(\frac{1}{e^{1/2}}\right)^{n-1} $$

However I first need to prove that $$\frac{1}{e^{k/2}} \geq \frac{k}{e^k} $$

and I get stuck at $$e^{k/2} \geq k$$

However I know that both of the derivatives of the respectives functions on real numbers is positive and the growth rate of $e^k$ is greater than $k$ but I doubt that its convincing enough.

Thank you for your help!

Answer 1

Let prove by induction

BASE CASE

$k=1 \implies e>1$

INDUCTION STEP

Suppose $e^k>k$ we want to prove that $e^{k+1}>k+1$

$e^{k+1}=e\cdot e^k\stackrel{\text{inductive hypothesis}}>e\cdot k>2\cdot k=k+k>k+1\quad \square$

Answer 2

You can use induction on $k\in\Bbb N_{\ge 0}$. You already knows that $e^0=1>0$, so the base case holds. Now assume that $e^k>k$ and you want to prove that $e^{k+1}>k+1$, what is straightforward because

$$e^{k+1}=e^k \cdot e>e^k\cdot 2=e^k+e^k>e^k+1$$ for all $k\in\Bbb N_{\ge 0}$.

---

However showing that $e^{k/2}>k$ by induction is not as easy. We want to show the equivalent inequality $e^k>k^2$, and we choose this time $k=3$ as the base case, what holds because

$$e^3>\left(\frac{5}2\right)^3=\frac{125}{8}>\frac{80}8=10>3^2$$

Assuming the hypothesis of $e^k>k^2$ we want to show that $e^{k+1}>(k+1)^2$. Now observe that

$$e^k\ge 2k+1 \implies e^{k+1}>e^k+e^k>k^2+2k+1=(k+1)^2$$

Then if we shows that $e^k\ge 2k+1$ for all $k\ge 3$ we are done, so we will do a second induction to prove this inequality. Clearly the base case $k=3$ holds because $e^3>2^3=8> 2\cdot 3+1=7$. Assuming the hypothesis $e^k\ge 2k+1$ now we want to show that $e^{k+1}\ge 2(k+1)+1$ what is true because

$$e^{k+1}>2e^k\ge 2k+3\implies e^k\ge k+\frac32$$

and $2k+1>k+\frac32$ holds for all $k\ge 1$.

Thus we have proved that $e^k\ge 2k+1$ for all $k\ge 3$, and consequently that $e^k>k^2$ for all $k\ge 3$. By last observe that the inequality $e^k>k^2$ also holds for $k=0,1,2$, thus we finally can conclude that $e^{k/2}>k$ for all $k\in\Bbb N_{\ge 0}$, as desired.

Answer 3

As $e^k > 0$ we have $e^k > k$ for all $k \leq 0$.

The Taylor series of the exponential converges for all values of $k \in \mathbb{R}$. Thus when $k$ is positive

\begin{align*} e^k & = 1 + k + \frac{k^2}{2!}+\frac{k^3}{3!}+ ... \\ & > k \end{align*}

as the terms $1 + \frac{k^2}{2!} + \frac{k^3}{3!} + ... $ are strictly positive for $k>0$.

Answer 4

The exponential is its own derivative, hence by the Mean Value Theorem, $$ e^x-e^0=(x-0)\cdot e^\xi$$ for some $\xi$ between $0$ and $x$, i.e., $$\tag1 e^x=1+xe^\xi.$$ We also know that $e^\xi>1$ for $\xi >0$ and $e^\xi<1$ for $\xi<0$. This makes $xe^\xi\ge x$, so that $(1)$ leads to $$ e^x\ge 1+x\qquad \text{for all }x\in\Bbb R.$$

By the multiplicativity of the exponential, we conclude further that $$e^x=(e^{x/2})^2 \ge (1+x)^2\qquad\text{for all }x\ge -1.$$

Answer 5

If we use the power series of $e^x$ that is

$e^x=\sum_{k=0}^{\infty}\frac{x^n}{n!},$

we get for $x\geq0$

$e^{x}=\sum_{k=0}^{\infty}\frac{x^n}{n!}\geq \frac{x^1}{1!}=x.$

The case $x<0$ is clear :-)

Answer 6

If you use that $e^t\ge 1$ for all $t\ge 0$, then the fundamental theorem of calculus shows that $e^k-1=\int_0^k e^tdt=k\int_0^1e^{kt}dt\ge k$, so $e^k\ge 1+k>k$ for all $k\ge 0$.

Answer 7

$f(k)=e^k-k$

$f'(k)=e^k-1=0\iff k=0$, and $f''(k)=e^k>0$ for all $k$. Therefore it $f$ has a global minimum at $k=0$, where $f(0)=0$, so $f(k)\geq 0$ for all $k$.

Answer 8

Observe that $$ e^k=(1+\zeta)^k\geq1+k\zeta>1+k>k $$ by the binomial theorem where $\zeta=e-1>1$ (this can be obtained by looking at the series definition of $e$ for example).

Note that in using the comparison test we only need that $e^k>k$ for sufficiently large $k$ and this can be proven by considering $\lim_{k\to\infty}\frac{k}{e^k}$. Note that $$ 0\leq\frac{k}{e^k}\leq\frac{k}{2^k}\leq\frac{k}{k^2}=\frac{1}{k} \quad (k\geq4) $$ since $e\ge2$ and $2^k\geq k^2$ for $k\geq 4$ (can be proven by induction). Thus $$ \frac{k}{e^k}\to 0 $$ In particular, (by the definition of the limit) there exists $N$ such that $n\geq N$ implies that $\frac{k}{e^k}<1$ as desired.

Answer 9

$f(x)=e^x$ is a convex function since $f(x)>0$ and $f''(x)=f(x)$. In particular the graph of $f(x)$ lies above the tangent at the origin and $e^x\geq x+1$ holds for any $x\in\mathbb{R}$. $e^k>k$ is a trivial consequence.