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 !
Best Answer
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 $o(\ln n)$.