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.