Where does $e$ come in the picture??!?!??

Can someone please help me by proving:

$\frac { 1 }{ 1! } +\frac { 1 }{ 2! } +\frac { 1 }{ 3! } +\frac { 1 }{ 4! } +.....\quad =\quad e-1$

Thanks!!

#Algebra

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Comments

Use the infinite series expansion of $e^x$ where $x=1$ then subtracting by $1$ will give you the LHS.

Marc Vince Casimiro - 6 years, 5 months ago

Can be done easily using expansion of e to the power x

Abhijeet Verma - 6 years, 5 months ago

Can you please elaborate?

A Former Brilliant Member - 6 years, 5 months ago

e^x=1+x+x^2/2!+x^3/3!+.... this implies e^x-1=x+x^2/2!+x^3/3!+.... putting x=1, e-1=1+1/2!+1/3!+....

Abhijeet Verma - 6 years, 5 months ago

it is clear from general form.. e^x =1+x+(x^2)/2! +(x^3)/3!+ (x^4)/4!........................---(1)

subtracting 1 from both sides (e^x)-1=x+(x^2)/2! +(x^3)/3!+ (x^4)/4!........................---(2)

comparing ur ques. and R.H.S. of eqn. (2) we get x=1 so ur ans. is e-1 regards-Arsh Sekhon........... hope you'll learn it at edusquare in limits and derivatives soon..........

Arsh Sekhon - 6 years, 5 months ago

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Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Where does eee come in the picture??!?!??

Comments

Where does $e$ come in the picture??!?!??