$2 < (1 + \frac{1}{n})^n < 3$ for all $n > 2$ Show by induction that $2 < (1 + \frac{1}{n})^n < 3$ for all natural n > 2.


*

*Induction base. For $n = 3$ the inequality is obviously  true.

*Assume that for $n = k$ inequality is true, than i can prove $(1 + \frac{1}{n+1})^{n +1} < 3$.
$$\frac{(n + 1)^{n}}{n^{n}} < 3 \\
1 < \frac{3n^{n}}{(n+1)^{n}} \\
\frac{(n+2)^{n+1}}{(n+1)^{n+1}} < \frac{3 n^{n} (n + 2)^{n+1} }{(n+1)^{2n +1}}, but \\
\frac{3 n^{n} (n + 2)^{n+1} }{(n+1)^{2n +1}} < 3, then \\
\frac{(n+2)^{n+1}}{(n+1)^{n+1}} < 3
$$
How can i prove, that $(1 + \frac{1}{n+1})^{n +1} > 2$
 A: HINT
What about Bernoulli's inequality?
A: By the binomial theorem, we have
$$
\left(1+\frac1n\right)^n = 1 + \binom n1\cdot \frac1n + \cdots
$$
where the first two terms are both $1$, and the rest ($n>2$ means there is at least one more term) are strictly positive.
In fact, you can show that $(1+\frac1n)^n<3$ with the same binomial theorem, noting that for $i\geq 1$, we have
$$
\binom ni\frac1{n^i} <\frac1{2^{i-1}}\tag{*}
$$
Using this, we get
$$
\left(1+\frac1n\right)^n = 1 + \sum_{i = 1}^n\binom ni\frac1{n^i}<1 + \sum_{i = 1}^n\frac1{2^{i-1}} = 1+2 = 3
$$

To prove $\text{(*)}$, I will show the left inequality of
$$
\binom ni2^{i-1}\leq n\cdot(n-1)\cdots(n-i+1)<n^i
$$
Say we have a game we can play which takes $i$ players, divided into two (not necessarily equally sized) teams. Then the left-hand side shows counts how many starting match-ups you can make from a group of $n$ people by first picking $i$ players, then dividing those $i$ players into two teams.
On the other hand, the right-hand side counts the number of ways I can pick $i$ of the $n$ players and line them up in a row. Assuming none of them have the same name, if we say everyone ahead of the person with the first name (alphabetically) in that line is on one team, and then that person and every person behind him is on the other team, then each possible way to divide up teams for the game may be realized by lining up the players like this. Thus the number of lines is at least the same as the number of team compositions, proving the inequality.
A: I'd like to point out an error in the proof for $\left(1+\frac1n\right)^n < 3$.
The step that says $\frac{3 n^{n} (n + 2)^{n+1} }{(n+1)^{2n +1}} < 3$ is wrong (I plugged $n=3$ into the windows calculator and got a term slightly above 3). That term is equal to
$$\frac{3(n^2 + 2n)^n(n+2)}{(n^2+2n+1)^n(n+1)} = 3\frac{(n^2 + 2n)^n}{(n^2+2n+1)^n}\frac{n+2}{n+1}$$
so the product of $3$, a term sligthly below $1$ and a term slightly above $1$. I suspect it will be $>3$ for all cases, but that might not be easy to proof.
