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# 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!!

Note by Abhineet Nayyar
2 years ago

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Use the infinite series expansion of $$e^x$$ where $$x=1$$ then subtracting by $$1$$ will give you the LHS. · 2 years ago

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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.......... · 2 years ago

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Can be done easily using expansion of e to the power x · 2 years ago

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Can you please elaborate? · 2 years ago

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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!+.... · 2 years ago

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