Lower bound for $\sqrt{x} - \sqrt{x-1} $

Question

Does anyone know any useful lower bound for $\sqrt{x} - \sqrt{x-1} $ for $x>1$. I have a problem where I want to find a lower bound for $$\sqrt{C \log(n)} - \sqrt{C \log(n)-1} $$ for a positive constant C and $n$ large enough.

Edit: Sorry for not being precise, a lower bound as a function of $x$.

Answer 1

As noted in a comment,

$\begin{array}\\ \sqrt{x} - \sqrt{x-1} &=(\sqrt{x} - \sqrt{x-1})\dfrac{\sqrt{x} + \sqrt{x-1} }{\sqrt{x} + \sqrt{x-1} }\\ &=\dfrac1{\sqrt{x} + \sqrt{x-1} }\\ &\gt \dfrac1{2\sqrt{x}}\\ \end{array} $

Using the generalized binomial theorem,

$\begin{array}\\ \sqrt{1-x} &=\sum_{n=0}^{\infty} \binom{\frac12}{n}(-x)^n\\ &=1+\sum_{n=1}^{\infty} \dfrac{\prod_{k=0}^{n-1}(\frac12-k)}{n!}(-1)^nx^n\\ &=1+\sum_{n=1}^{\infty} \dfrac{\prod_{k=0}^{n-1}(1-2k)}{2^nn!}(-1)^nx^n\\ &=1+\sum_{n=1}^{\infty} \dfrac{\prod_{k=0}^{n-1}(2k-1)}{2^nn!}x^n\\ &=1-\sum_{n=1}^{\infty} \dfrac{\prod_{k=1}^{n-1}(2k-1)}{2^nn!}x^n\\ &=1-\sum_{n=1}^{\infty} \dfrac{\prod_{k=1}^{n-1}(2k-1)\prod_{k=1}^{n-1}(2k)}{2^nn!\prod_{k=1}^{n-1}(2k)}x^n\\ &=1-\sum_{n=1}^{\infty} \dfrac{(2n-2)!}{2^nn!2^{n-1}(n-1)!}x^n\\ &=1-\sum_{n=1}^{\infty} \dfrac{(2n-2)!}{2^{2n-1}n!(n-1)!}x^n\\ &=1-\sum_{n=1}^{\infty} \binom{2n-2}{n-1}\dfrac{1}{n2^{2n-1}}x^n\\ &=1-\frac12 x-\frac18 x^2-\frac1{16}x^3-...\\ \end{array} $

so

$\begin{array}\\ \sqrt{x}-\sqrt{x-1} &=\sqrt{x}(1-\sqrt{1-\frac1{x}})\\ &=\sqrt{x}(\frac1{2 x}+\frac1{8 x^2}+\frac1{16x^3}+...)\\ &=\frac1{2 \sqrt{x}}+\frac1{8 x\sqrt{x}}+\frac1{16x^2\sqrt{x}}+...\\ &=\frac1{\sqrt{x}}(\frac1{2}+\frac1{8 x}+\frac1{16x^2}+...)\\ \end{array} $

You can stop the series at any desired term to get a lower bound.

Answer 2

Hmm. $$\sqrt x-\sqrt{x-1}=\sqrt x\left(1-\sqrt{1-1/x}\right).$$ It's easy to see with a little calculus (Mean Value Theorem) that $$1-\sqrt{1-t}\ge ct\quad(t\to0)$$and there you are.

Answer 3

There would be many ways to bound this, for e.g. recognising that this can be written as the integral of a convex function, the rather well known Hermite–Hadamard inequality gives

$$\sqrt{x}-\sqrt{x-1}=\int_{x-1}^x\frac{dt}{2\sqrt{t}} \geqslant \frac1{\sqrt{4x-2}}$$

For deciding which bound is better to use, we need more criteria/ conditions.