Prove that $2\lg(n+1)-2\le 2\lg n,$ given $n\ge 1$

Question

I want to prove

$$2\lg(n+1)-2\le2\lg n, \text{given } n\ge1.$$

Since that $2\lg(n+1)-2 = 2(\lg(n+1)-1)$, so this is equivalent to prove $\lg(n+1)-1\le\lg n.$ And $\lg(n+1)-1 = \lg(n+1)-\lg 2 = \lg(\large\frac{n+1}{2})$ so this is equivalent to prove ${\large\frac{n+1}{2}} \le n,$ given $n\ge 1.$ Finally, since

$$\frac{n+1}{2}\le n\iff 1\le n.$$

And this does hold given $n\ge 1$.

Is my proof correct? I'm thinking whether this can be proved in one line.

Answer 1

Your proof is correct. I would write $$ \lg (n+1) \le \lg (n+n) = \lg (2n) = \lg2 + \lg n = 1 + \lg n $$ which holds for $n \ge 1$ (with equality exactly for $n=1$). Then a simple rearrangement and multiplication with $2$ gives the desired estimate.