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For an positive integer $n\ge 2$, is there a good upper bound for $(x-1)^n-x^n$ for $x> 1$?
Revised question: Is it less than $-\log(x-1)$ for all $x>1$?

By Binomial theorem, $(x-1)^n-x^n=\sum_{k=1}^n\binom{n}{k}x^{n-k}(-1)^k$. Or by Mean Value Theorem, it is equal to $-n\xi^{n-1}$ for some $\xi\in(x-1,x)$. I wonder if there are good estimate of this function?

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  • $\begingroup$ Maybe I'm not understanding but isn't the maximum for that function with a positive integer $n\ge 2$ always $-1$ with $x=1$ ? $\endgroup$
    – JimB
    Nov 9, 2017 at 3:06
  • $\begingroup$ @JimB That upper bound is not good for large $x$. $\endgroup$
    – kccu
    Nov 9, 2017 at 3:09
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    $\begingroup$ what are you trying to achieve? In a way you've just made your expression more complicated. Calculating $(x-1)^n-x^n$ is a lot simpler than the sum. $\endgroup$
    – videlity
    Nov 9, 2017 at 3:13
  • $\begingroup$ I wonder if it less than $-\log(x-1)$? $\endgroup$
    – Connor
    Nov 9, 2017 at 3:16
  • $\begingroup$ @kccu. Then I'm certainly misinterpreting the problem. I was interpreting it as fixing $n$ and then finding the value of $x$ that maximizes the function. How large is large? (x - 1)^n - x^n /. {n -> 5, x -> 100000000} results in -499999990000000099999999500000001 using Mathematica. $\endgroup$
    – JimB
    Nov 9, 2017 at 3:17

2 Answers 2

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The upper bound $-n\zeta^{n-1} \le -n(x-1)^{n-1}$ is certainly less that $-\log(x-1)$ for $x \ge 2.

You can get better upper bounds by using Taylor approximations of $f(t) = t^n$ about $t = x$ and Lagrange's remainder term. For example: $$ (x-1)^n-x^n = -nx^{n-1} + \frac{1}{2} n(n-1)\zeta^{n-2} \le -nx^{n-1} + \frac{1}{2} n(n-1)x^{n-2} \, . $$ and $$ (x-1)^n-x^n = -nx^{n-1} + \frac{1}{2} n(n-1)x^{n-2} - \binom{n}{3} \zeta^{n-3} \le -nx^{n-1} + \frac{1}{2} n(n-1)x^{n-2} - \binom{n}{3}(x-1)^3\, . $$

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Edit: Made a mistake and the question changed.

Claim: $(x-1)^n-x^n < -\log(x-1)$ for $x>1$.

This is equivalent to showing that

$$ (x+1)^n-x^n > \log x, \qquad x>0 $$ Now \begin{align}(x+1)^n-x^n &= nx^{n-1} + {n\choose 2} x^{n-2} + .... \\&>nx^{n-1} \\&> x \qquad {\because n\geq 2 } \\&>\log x. \end{align}

As mentioned in the other answer, this bound is quite poor.

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    $\begingroup$ That cannot be correct. Compare the polynomial degrees of the two sides. $\endgroup$ Nov 9, 2017 at 3:30
  • $\begingroup$ I just revised the question. I wonder if it is less than $-\log(x-1)$ for all $x>1$? It is definitely true for $x$ large enough by your estimate. But what about small $x$? $\endgroup$
    – Connor
    Nov 9, 2017 at 3:30
  • $\begingroup$ It is true. I will add a proof in an answer. $\endgroup$
    – videlity
    Nov 9, 2017 at 3:33
  • $\begingroup$ @HansEngler you are right. I made a mistake. $\endgroup$
    – videlity
    Nov 9, 2017 at 3:36
  • $\begingroup$ @Connor updated the answer. $\endgroup$
    – videlity
    Nov 9, 2017 at 3:43

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