Find asymptotically equivalent function

Question

I'm trying to understand the steps of this example (finding an asymptotically equivalent function). But the last two stay a little foggy for me.

$(n+1)\ln (n+1)-n\ln n$

$=(n+1)\ln\left(1+\frac{1}{n}\right)+\ln n$

$=(n+1)\left(\frac{1}{n}+o\left(\frac{1}{n}\right)\right)+\ln n$

$=1+\frac{1}{n}+o(1)+\ln n$

$=\ln n+o(\ln n)$

Hence $(n+1)\ln (n+1)-n\ln n \sim \ln n$

Why can we get rid of the $\frac{1}{n} + o\left(\frac{1}{n}\right)$ ?

And how comes the $o(\ln n)$ ?

More details and explanations would be welcome !

Answer 1

If I understand your difficulty, here are more intermediate steps: $$(n+1)\biggl(\frac{1}{n}+o\Bigl(\frac{1}{n}\Bigr)\biggr)=\frac{n+1}n+o\Bigl(\frac{n+1}{n}\Bigr)=1+\frac 1n+o(1)$$ since $\frac{n+1}n\sim_\infty 1$ and hence $\:o\bigl(\frac{n+1}{n}\bigr)=o(1)$.

Last step: the constant $1$ is $o(\ln n)$, hence $o(1)$ is too, $\frac1n=o(\ln n)$, so their sum is also $o(\ln n)$.

Answer 2

As a check, since $(x\ln(x))' =\ln(x)+1 $, by the mean value theorem, $(n+1)\ln(n+1)-n\ln(n) =\ln(n+c)+1 $ where $0 \le c \le 1$.

Then $\ln(n+c)-\ln(n) =\ln(1+c/n)$. Since $0 \le \ln(1+x) \le x$ for $x \ge 0$, $\ln(1+c/n) =ac/n $ where $0 \le a \le 1$, so $(n+1)\ln(n+1)-n\ln(n) =\ln(n+c)+1 =\ln(n)+1+b/n $ where $0 \le b \le 1$.