Proving e^x > sigma(x^i/i!) for every x>0 | Induction Method

[madah12]

Homework Statement

prove that e^x > sigma from i= o to n (x^i/i!) for every x>0

Homework Equations

The Attempt at a Solution

I will do it by induction
for n=1
e^x > x+1
but e^0=1 and 0+1=1
f(x)=e^x , g(x)=x+1
f(0)=g(0)
f'(x)=e^x , g'(x)=1
e^x>1 for every x >0
so f(x)>g(x)
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now assuming it's true for n =k
e^x> sigma from i= o to k (x^i/i!) for every x>0 (a)
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we prove it is true for k+1
h(x)=sigma from i= o to k+1 (x^i/i!) for every x>0
f(0)=h(0)=1
f'(x)=e^x , h'(x) = sigma from i= o to k (x^i/i!)
and by our assumption e^x >sigma from i= o to k (x^i/i!)
therefore f'(x)>h'(x) for every x >0
f(x)>h(x)
<=>e^x >sigma from i= o to k+1 (x^i/i!) for
and since it is proven for n=k+1
then by induction
e^x >sigma from i= o to n (x^i/i!)

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is this right? and how can I prove it is true if n --> infinity?

 

[LCKurtz]

Your argument looks ok to me. So you have

\sum_{k=0}^n\frac {x^k}{k!} &lt; e^x

for all n > 0, x>0. If you wish to take the limit as n→ ∞, you must admit the possibility of equality:

\sum_{k=0}^\infty\frac {x^k}{k!} \le e^x

You know the partial sums converge since they are increasing and bounded above by e^x. Less than or equal is all this argument shows, although when you study Taylor series you will learn that equality holds in the limit.