Question

The lowest integer which is greater than \(\left(1+\frac{1}{10^{100}}\right)^{10^{100}}\) is ______________.

• A. 3
• B. 4
• C. 2
• D. 1

Hint:

Expand binomially: the value lies strictly between 2 and \(e\lt 3\).

Answer 1

The Correct Option is A

Let \(n=10^{100}\). Then \(\left(1+\frac1n\right)^n=\sum_{k=0}^{n}\binom nk\frac1{n^k}=1+1+\frac{1-\frac1n}{2!}+\frac{(1-\frac1n)(1-\frac2n)}{3!}+\dots\)
All terms after the first two are positive, so the value is greater than 2.
Also \(\binom nk\frac1{n^k}\lt\frac1{k!}\), so the value is less than \(\sum_{k\ge0}\frac1{k!}=e\lt3\).
Thus \(2\lt\left(1+\frac1n\right)^n\lt3\) and the lowest integer greater than it is 3, which is option (1).
Answer: (1) 3