The lowest integer which is greater than (1+110^100)^10^100 is

Question

The lowest integer which is greater than $\left(1+\dfrac{1}{10^{100}}\right)^{10^{100}}$ is

Accepted Answer

Let $$P = {\left( {1 + {1 \over {{{10}^{100}}}}} \right)^{{{10}^{100}}}}$$

Let $x = {10^{100}}$

$$ \Rightarrow P = {\left( {1 + {1 \over x}} \right)^x}$$

$$ \Rightarrow P = 1 + (x)\left( {{1 \over x}} \right) + {{(x)(x - 1)} \over {\left| \!{\underline {\,
  2 \,}} \right. }}.{1 \over {{x^2}}} + {{(x)(x - 1)(x - 2)} \over {\left| \!{\underline {\,
  3 \,}} \right. }}.{1 \over {{x^3}}} + ....$$

(upto 10100 + 1 terms)

$$ \Rightarrow P = 1 + 1 + \left( {{1 \over {\left| \!{\underline {\,
  2 \,}} \right. }} - {1 \over {\left| \!{\underline {\,
  2 \,}} \right. {x^2}}}} \right) + \left( {{1 \over {\left| \!{\underline {\,
  3 \,}} \right. }} - ...} \right) + ...$$ so on

$$ \Rightarrow P = 2 + \left( {Positive\,value\,less\,than\,{1 \over {\left| \!{\underline {\,
  2 \,}} \right. }} + {1 \over {\left| \!{\underline {\,
  3 \,}} \right. }} + {1 \over {\left| \!{\underline {\,
  4 \,}} \right. }} + ...} \right)$$

Also $$e = 1 + {1 \over {\left| \!{\underline {\,
  1 \,}} \right. }} + {1 \over {\left| \!{\underline {\,
  2 \,}} \right. }} + {1 \over {\left| \!{\underline {\,
  3 \,}} \right. }} + {1 \over {\left| \!{\underline {\,
  4 \,}} \right. }} + ...$$

$$ \Rightarrow {1 \over {\left| \!{\underline {\,
  2 \,}} \right. }} + {1 \over {\left| \!{\underline {\,
  3 \,}} \right. }} + {1 \over {\left| \!{\underline {\,
  4 \,}} \right. }} + ... = e - 2$$

$$\Rightarrow$$ P = 2 + (positive value less than e - 2)

$$\Rightarrow$$ P $$\in$$ (2, 3)

$$\Rightarrow$$ least integer value of P is 3

Solutions · 1