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If $t_n=\bigg(1+\large\frac{1}{n}\bigg)$$^n$ for $n\in N$, then $t_n$ lies in the interval?

$\begin{array}{1 1} 1 \leq t_n < 2 \\ 2 \leq t_n 3 \\ 2 < t_n \leq 3 \\ t_n > 3 \end{array} $

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Toolbox:
  • $1+x+x^2+x^3+..........=\large\frac{1}{1-x}$ if $|x|<1$
$t_n=\bigg(1+\large\frac{1}{n}$$\Bigg)^n=^nC_0+^nC_1\large\frac{1}{n}+$$^nC_2(\large\frac{1}{n})$$^2+........$
$=1+1+\large\frac{n(n-1)}{2!}.(\frac{1}{n})^2+.......$
$=2+\large\frac{(1-\frac{1}{n})}{2!}$$+\large\frac{(1-\frac{1}{n})(1-\frac{2}{n})}{3!}+........$
$=2+S$ (say) where $S$ is non negative real number.
$S=\large\frac{(1-\frac{1}{n})}{2!}$$+\large\frac{(1-\frac{1}{n})(1-\frac{2}{n})}{3!}+........$
$\Rightarrow\:S<\large\frac{1}{2!}+\frac{1}{3!}+.....$
$\Rightarrow\:S<\large\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...$$=\large\frac{\frac{1}{2}}{1-\large\frac{1}{2}}=1$
$\Rightarrow\:2\leq t_n<3$
answered Oct 8, 2013 by rvidyagovindarajan_1
 

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