Proving that log(n+1) is O(log(n). If true, find C and K

Question

If I am not mistaken $\log(n+1)$ is $O(\log(n))$ since $\lim_{x\to \infty}(\frac{\log(x+1)}{\log(x)}) = 1$

But I don't know how to find $C$ and $K$. Can someone show me how?

Thank you so much!

Answer 1

$\log (n+1)\leq C \log (n)$ iff $n+1 \leq n^{C}$. Since $n+1 \leq 2n \leq n^{2}$ for $n \geq 2$ we see that $\log (n+1)\leq 2\log (n)$ for $n \geq 2$.

Answer 2

$\ln(n+1)-\ln(n) =\ln(1+1/n) \lt 1/n$ so $\ln(n+1) =\ln(n)+O(1/n) =O(\ln(n)) $ since $\ln(n) = o(n) $.