Proving Inequality with Bernoulli's: k≤n Positive Integers

[bennyzadir]

Let be $$k \leq n$$ poitive integers. How to show that
$$\left (1+\frac1 n \right)^k \leq 1 + \frac{ke}{n}$$ .

It seems to me that it has something to do with Bernoulli's inequality.
Thank you in advance!

 

[Gib Z]

Are you allowed to use some basic results from calculus for this problem? Note that $$\left( 1 + \frac{1}{n} \right)^n$$ is monotonically increasing to e, so $$\left( 1 + \frac{1}{n} \right)^k = \left(\left( 1 + \frac{1}{n} \right)^n\right)^{\frac{k}{n}} < e^{\frac{k}{n}}$$. Thus it is sufficient to show that $$e^{\frac{k}{n}} \leq 1 + \frac{ke}{n}$$. Let $a= k/n \leq 1$. Then rearranging the required inequality, we have to show $$\frac{ e^a - e^0 }{ a- 0} \leq e$$, which follows quite quickly from the Mean Value Theorem.

 

[bennyzadir]

Are you allowed to use some basic results from calculus for this problem? Note that $$\left( 1 + \frac{1}{n} \right)^n$$ is monotonically increasing to e, so $$\left( 1 + \frac{1}{n} \right)^k = \left(\left( 1 + \frac{1}{n} \right)^n\right)^{\frac{k}{n}} < e^{\frac{k}{n}}$$. Thus it is sufficient to show that $$e^{\frac{k}{n}} \leq 1 + \frac{ke}{n}$$. Let $a= k/n \leq 1$. Then rearranging the required inequality, we have to show $$\frac{ e^a - e^0 }{ a- 0} \leq e$$, which follows quite quickly from the Mean Value Theorem.

Thank you very much for your clear and understandable answer.