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# If $t_n=\bigg(1+\large\frac{1}{n}\bigg)$$^n for n\in N, then t_n lies in the interval? \begin{array}{1 1} 1 \leq t_n < 2 \\ 2 \leq t_n 3 \\ 2 < t_n \leq 3 \\ t_n > 3 \end{array} Can you answer this question? ## 1 Answer 0 votes Toolbox: • 1+x+x^2+x^3+..........=\large\frac{1}{1-x} if |x|<1 t_n=\bigg(1+\large\frac{1}{n}$$\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!}+........$
$\Rightarrow\:S<\large\frac{1}{2!}+\frac{1}{3!}+.....$