1 Answer

Abhik Bose Pramanik · Answer 1 · 2021-07-18T21:34:50+0000

FYI, that L in the denominator is factorial.

This is some pretty serious stuff for high school.

My inclination is that the answer is A, which is the definition of e. Let's see if we can show this.

Let's write

$\displaystyle f(n) = \left(1 + \frac 1 n \right)^n$

Let's expand this with the binomial expansion

$\displaystyle f(n) =\sum_(k=0)^n {n \choose k} (1)/(n^k)$

$\displaystyle f(n) =\sum_(k=0)^n (n!)/(k!(n-k)!) \cdot (1)/(n^k)$

$\displaystyle f(n) =\sum_(k=0)^n (n(n-1)\cdots(n-k+1))/(k! \, n^k)$

Let's focus on when n is really big and on the ks that are relatively small, which make up the bulk of the sum as the terms get small rapidly.

Then that numerator n(n-1)···(n-k+1) ≈ n^k as all the factors are about n.

$\displaystyle f(n)\approx \sum_(k=0)^n (n^k)/(k!\, n^k)$

$\displaystyle f(n)\approx \sum_(k=0)^n (1)/(k!)$

OK, we showed for large n this is approximately true, and it will be exactly true in the limit, so we choose

Answer: A

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