Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

It was shown in here that $\left(1+\frac{1}{n}\right)^n < n$ for $n>3$. I think we can be come up with a better bound, as follows:

$$\left(1+\frac{1}{n}\right)^n \le 3-\frac{1}{n}$$ for all natural number $n$.

The result is true for all real number $\ge 1$, which can be shown using calculus. I wonder if the above result can be proved using mathematical induction?

I have tried but fail! Anyway, this question is also inspired by, and related to this question.

Edit:

I also found that $$\left(1+\frac{1}{n+k}\right)^n \le 3-\frac{k+1}{n}$$ for all natural number $k$, some large $N$ and $n > N$. This implies that $$\left(1+\frac{1}{2n}\right)^n \le 2-\frac{1}{n}.$$

And again, I can't prove any of them using Mathematical Induction.

share|improve this question
    
In order to use induction, I need to have $\left(1+\frac{1}{n+1}\right)^n < 3-\frac{2}{n}$ for $n\ge 2$, which again is not trivial. Equivalently, it is to prove $\left(1+\frac{1}{n}\right)^n -\left(1+\frac{1}{n+1}\right)^n >\frac{1}{n}$ for $n \ge 2$. –  pipi Feb 20 '13 at 0:53

1 Answer 1

up vote 14 down vote accepted

I prove this inequality does not use induction, but I think this proof also elementary proof, because this proof does not use calculus.

$$\begin{array}{lcl} \left( 1+\frac{1}{n}\right)^n &=& 1+\binom{n}{1} \frac{1}{n}+ \sum_{k=2}^n\binom{n}{k} \frac{1}{n^k}\\ &=&2+\sum_{k=2}^n \frac{1}{k!} \frac{n(n-1)\cdots (n-k+1)}{n^k}\\ &\le& 2+\sum_{k=2}^n \frac{1}{k!} \le 2+\sum_{k=2}^n \frac{1}{k(k-1)} \\ &=& 3-\frac{1}{n} \end{array} $$

share|improve this answer
    
Typo, 1st line, right side, $n$ should be $1/n$. –  Gerry Myerson Feb 19 '13 at 4:52
2  
Nice proof. +1. –  user1551 Feb 19 '13 at 4:55
    
@GerryMyerson Thanks! –  tetori Feb 19 '13 at 4:56
    
@tetori, thanks for the nice proof. Anyway, I am thinking of mathematical induction... –  pipi Feb 19 '13 at 5:26

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.