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Prove that $2 < f(n) < 3$ if $n$ is a natural number and $n \ge 2$ and $f(n)=\left (1+\frac{1}{n}\right)^n$.

I've succeded in proving that $f(n)>2$ :

$$ f(n)=\left (1+\frac{1}{n}\right)^n \\ f(n)=\binom{n}{0}+\binom{n}{1}\left(\frac{1}{n}\right)+\binom{n}{2}\left(\frac{1}{n^2}\right)+\binom{n}{3}\left(\frac{1}{n^3}\right)+...+\binom{n}{n}\left(\frac{1}{n^n}\right) $$ Since the first two terms in this series are $1$, $f(n) > 2$. Can anybody give me a few hints on proving the upper bound ? Edit : I already know of the way to do this using limits and am looking for a way to do this using the Binomial Theorem.

Vishnu V.S
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    Use the fact that $$\lim_{n\to \infty} (1+\frac {1}{n})^n =e $$ –  Jan 18 '17 at 15:46
  • @Rohan Thanks for the answer but I'm looking for a way to do this without limits as this is an exercise in the chapter - " Binomial Theorem " in one of my father's old math textbooks. – Vishnu V.S Jan 18 '17 at 15:49
  • See my answer http://math.stackexchange.com/questions/489606/how-can-one-prove-that-e-pi/489638#489638 for a hint of a way to proceed – Mark Bennet Jan 18 '17 at 15:50
  • @MarkBennet This still requires us to use limits. Right ? – Vishnu V.S Jan 18 '17 at 15:52
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    @VishnuV.S The method of bounding the terms individually can be used to compare with a finite geometric progression. – Mark Bennet Jan 18 '17 at 15:54
  • @MarkBennet The terms, apart from the first two are successively decreasing. But how do we prove that their sum is less than 1 ? As for a geometric progression, I can't figure out a common ratio ? – Vishnu V.S Jan 18 '17 at 16:02
  • Try one half :-) – Simply Beautiful Art Jan 18 '17 at 16:11

1 Answers1

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By the AM-GM inequality, the sequence $\{a_n\}_{n\geq 1}$ given by $a_n=\left(1+\frac{1}{n}\right)^n$ is increasing:

$$ a_n = 1\cdot\left(1+\frac{1}{n}\right)^n \leq \left(\frac{1+n\cdot\left(1+\frac{1}{n}\right)}{n+1}\right)^{n+1} = a_{n+1} $$ but such sequence is also bounded above: $$ a_n = 2+\sum_{k=2}^{n}\binom{n}{k}\frac{1}{n^k} =2+\sum_{k=2}^{n}\frac{1}{k!}\prod_{j=0}^{k-1}\frac{n-j}{n}\leq 2+\sum_{k\geq 2}\frac{1}{k!}$$ and it is enough to prove that $\sum_{k\geq 2}\frac{1}{k!}<1$, that simply follows from: $$ \sum_{k \geq 2}\frac{1}{k!}<\sum_{k\geq 2}\frac{1}{k(k-1)}=\sum_{k\geq 2}\left(\frac{1}{k-1}-\frac{1}{k}\right)=1.$$

Jack D'Aurizio
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