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Define the following integral with $n$ an integer greater than $1$:

$$I_{n}=\int_{0}^{1}\frac{e^t}{(1+t)^n}dt.$$

Is it true that for all $n \geq 2$,

$$ \frac{1}{n-1}\left(1-\frac{1}{2^{n-1}}\right)\leq I_{n} \leq \frac{e}{n-1}\left(1-\frac{1}{2^{n-1}}\right)?$$

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You can compute this integral explicitly wolframalpha.com/input/… and then check inequalities. –  no identity Dec 31 '11 at 14:01
    
Hum ; Wolfram gives something with the incomplete Gamma function. Can I check any inequality with that ? –  user20010 Dec 31 '11 at 14:17

1 Answer 1

up vote 5 down vote accepted

We have $$\int_0^1\frac{dt}{(1+t)^n}=\left[\frac 1{1-n}(1+t)^{1-n})\right]_0^1=\frac {1-2^{1-n}}{n-1}=\frac 1{n-1}\left(1-\frac 1{2^{n-1}}\right),$$ and you can conclude since for $0\leq t\leq 1$, $1\leq e^t\leq e$.

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