Important Result (Comparison) - Practice Questions & MCQ

Important Result (Comparison)

$
2 \leq\left(1+\frac{1}{n}\right)^n<3, \quad n \in \mathbb{N}
$
Proof:
Expand, $\left(1+\frac{1}{n}\right)^n$ using binomial theorem

$
\begin{aligned}
\left(1+\frac{1}{n}\right)^n & =1+n \frac{1}{n}+\frac{n(n-1)}{2!} \frac{1}{n^2}+\frac{n(n-1)(n-2)}{3!} \frac{1}{n^3}+\cdots+\frac{n(n-1)(n-2) \cdots[n-(n-1)]}{n!} \frac{1}{n^n} \\
& =1+1+\frac{1}{2!}\left(1-\frac{1}{n}\right)+\frac{1}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)+\cdots+\frac{1}{n!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right) \cdots\left(1-\frac{n-1}{n}\right) \\
& <1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!} \\
& <1+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{n-1}}=1+1 \frac{\left\{1-\left(\frac{1}{2}\right)^n\right\}}{1-\frac{1}{2}}=1+2\left\{1-\left(\frac{1}{2}\right)^n\right\}=3-\frac{1}{2^{n-1}}
\end{aligned}
$
Hence, from above

$
2 \leq\left(1+\frac{1}{n}\right)^n<3, \quad n \geq 1.
$