Proving: 1+1/2!+1/3!...+1/n! < 2[1-(1/2)^n]

[Bazman]

Hi,

I need to prove the following:

1+ \frac{ 1}{ 2!} + \frac{1 }{3!} +...+ \frac{ 1}{ n!} &lt; 2 \lbrack 1 - ( \frac{ 1}{2 } )^n \rbrack

From trying various example I'm fairly sure the relation holds but I can't seem to prove it algebraically?

Does the ineqaulity make a difference? Or can you behave pretty mcu as if it was an "=" ?

I tried simply doing 2[1-(1/2)^n] + 1/(n+1)! to try to get to 2[1-(1/2)^n+1]
but I can't seem to get very far?

Can anyone shed any light?

 

[murshid_islam]

2 \left( 1 - \left( \frac{ 1}{2 } \right) ^n \right) = 1 + \frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{2^{n-1}}

we also know that 2^n &lt; n!

therefore,
\frac{1}{(n+1)!} &lt; \frac{1}{2^{n+1}}

so,
2 \left( 1 - \left( \frac{ 1}{2 } \right) ^n \right) + \frac{1}{(n+1)!} &lt; 2 \left( 1 - \left( \frac{ 1}{2 } \right) ^n \right) + \frac{1}{2^{n+1}}

2 \left( 1 - \left( \frac{ 1}{2 } \right) ^n \right) + \frac{1}{(n+1)!} &lt; 2 \left( 1 - \left( \frac{ 1}{2 } \right) ^{n+1} \right)

 

[Bazman]

thanks Murshid!

Its crystal clear now