Proving the Integral Bound for I_n: 0 < I_n < 1/(n+1)

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SUMMARY

The integral \( I_n = \int_{0}^{1} x^{n} e^{-x} dx \) is proven to satisfy the bounds \( 0 < I_n < \frac{1}{n+1} \). The proof establishes that \( I_n \) is greater than zero through integration by parts and demonstrates that \( I_n \) is less than \( \frac{1}{n+1} \) by comparing \( x^n e^{-x} \) to \( x^n \) over the interval [0,1]. The discussion highlights the importance of recognizing simpler approaches rather than relying on complex methods such as inverse substitution or factorial representations.

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Homework Statement


Let I_{n} = \int_{0}^{1}x^{n}e^{-x}dx

Show that 0 < I_{n} < \frac{1}{n+1}

Homework Equations


n/a


The Attempt at a Solution


I have been trying to prove this for a long time, and so far I haven't gotten anywhere. I managed to get a reduction formula for it using integration by parts, and I can prove that it is greater than 0, but not that it is less than 1/(n+1). I have tried using induction too, but with no luck.
 
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x^n*e^(-x) is less than x^n on [0,1], isn't it?
 
Ah, I can't believe I didn't think of that! I'd tried a bunch of complicated methods, trying to use inverse substition and even writing the antiderivative in terms of a factorial and summation, but it was right in front of me the whole time... Thank you!
 
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