I have seen in a proof that for $a,b$ positive integers and for $b\geq a+1$ it is true that $$(1-1/a)^{b-1}\leq e^{-1}$$ but this is not clear to me by the exponential series and the obvious upper bound is $1$.
Upper bound of $(1-1/a)^{b-1}$
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inequality
exponential-function
integers
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1Basically it's because $(1-1/n)^n$ approaches $1/e$ from below. – 2017-02-01
1 Answers
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Hint Here is one way - note $e^x$ is convex and hence stays above its tangent at $x=0$, so $e^x \ge 1+x$. Now set $x =-\frac1a$ and simplify.