A calculus problem by Archit Boobna

This is just Taylor's Expansion.

$\displaystyle \sum_{n=0}^{\infty} \dfrac {x^n}{n!} = e^x$

In this case, $x = 1$ so the answer is $e \approx 2.718$ .

1/0!+1/1!+1/2!+1/3!+1/4!+1/5!+1/6!+1/7!=2.718253968

e^x = 1 + x + x/2! + x/3! + x/4! + ............................................ putting x = 1 in the above equation e = 1/0! + 1/1! + 1/2! + 1/3! + ...................................... therefore the answer e= 2.718

The answer for the given question is e

e= $\displaystyle \sum_{n=0}^\infty$ $\frac {1}{n!}$ = 2.7182818284 5904523536 028747....

$Taylor \ polynomial$