# Is there a good upper bound for $(x-1)^n-x^n$ for $x\ge 1$ and $n=2,3,4,...$?

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?

• Maybe I'm not understanding but isn't the maximum for that function with a positive integer $n\ge 2$ always $-1$ with $x=1$ ?
– JimB
Nov 9, 2017 at 3:06
• @JimB That upper bound is not good for large $x$.
– kccu
Nov 9, 2017 at 3:09
• 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. Nov 9, 2017 at 3:13
• I wonder if it less than $-\log(x-1)$? Nov 9, 2017 at 3:16
• @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.
– JimB
Nov 9, 2017 at 3:17

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 $$(x-1)^n-x^n = -nx^{n-1} + \frac{1}{2} n(n-1)x^{n-2} - \binom{n}{3} \zeta^{n-3} \le -nx^{n-1} + \frac{1}{2} n(n-1)x^{n-2} - \binom{n}{3}(x-1)^3\, .$$ 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. • That cannot be correct. Compare the polynomial degrees of the two sides. Nov 9, 2017 at 3:30 • 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\$? Nov 9, 2017 at 3:30