Prove inequality e^{x}>\frac{x^{n+1}}{n+1!}

Nicontio1

Nicontio1

Answered question

2022-01-14

Prove inequality ex>xn+1n+1!

Answer & Explanation

Durst37

Durst37

Beginner2022-01-15Added 37 answers

Step 1
You can use Taylor series of the exponential, whose radius of convergence is infinite.
ex=k=0+xkk!
Can you proceed?
Observe that due to the previous series, we can write the partial sum for k from 0 to n+2:
1+x+x22!++xnn!+xn+1(n+1)!+xn+2(n+2)!+
And this, being a sum of positive terms, is surely greater than xn+1(n+1)!
SlabydouluS62

SlabydouluS62

Skilled2022-01-16Added 52 answers

Step 1
The inequality ex>xn+1n+1!
can be proven by
ex=l=0xll!l=n+1xll!xn+1(n+1)!

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