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I need to find an upper bound for $\left (1-\dfrac{1}{t}\right )^x$, where $x<t$. It should be of the form $c\dfrac{x}{t}$. Any ideas?

commie trivial
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Liad
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    Please, try this one: https://en.wikipedia.org/wiki/Bernoulli%27s_inequality – Arnaldo Jul 20 '22 at 16:43
  • Bernoulli gives a lower bound, but I need an upper bound. – Liad Jul 20 '22 at 16:48
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    More restrictions? $x$ could be negative? $t$ could be ${} < 1$? – GEdgar Jul 20 '22 at 16:54
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    I'm surprised this question got 2 upvotes as it lacks a lot of details about $x$ and $t$ as pointed out by GEdgar. Is $t$ fixed for this and then $x$ is allowed to vary? Or is it considered as a two-variable function? The question as it stands is of poor quality in my eyes... – Adam Rubinson Jul 20 '22 at 17:06
  • If $t\ge1$ and $0\lt x\lt t$ then $(1-t^{-1})^x\lt1+xt^{-1}(e-1)$ is a crude upper bound, somewhat in the form you ask for (actually, the proof relies on $x\lt t$ but since the quantity decreases in $x$, you recover the result for all $x\gt0$) – FShrike Jul 20 '22 at 17:25

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