$(1+\frac{1}{n\log n})^n-1=O(\frac{1}{n})$. When I solved a problem, I could solve it if I assumed that $(1+\frac{1}{n\log n})^n-1=O(\frac{1}{n})$
I tried to prove it, but I failed.
Actually, I don't convince if it is true.
Is it correct? If so, how to solve it?
 A: Bernoulli's inequality gives 
$$\left(1+{1\over n\log(n)}\right)^n\geq 1+{1\over \log(n)}. $$
A: $$\lim_n (1\pm \frac{1}{n\log n})^n=\lim_n [(1 \pm \frac{1}{n\log n})^{n\log(n)}]^\frac{1}{\log(n)}=(e^{\pm 1})^0=1 $$
A: $(1+1/(n \log(n))^n$ $\to 1 $  as  $n\to\infty$.
So , $(1+1/(n \log(n))^n$ =$O(1)$$ $
Since the function doesn't converge to $0$ , we conclude that
$f(n)$$\neq O(1/n)$
A: To show in an elementary way that
your expression is actually
of order $1/\log n$,
I will use this "contra-Bernoulli" inequality:
If $n$ is a positive integer and $0 < x < 1/n$ then $(1+x)^n < 1/(1-nx)$.
This is readily proved by induction in the form
$(1-nx)(1+x)^n< 1$.
Putting $x = 1/(n \log n)$, this becomes
$(1+1/(n \log n))^n < 1/(1-n/(n \log n))
= 1/(1-1/\log n)
= \log n/((\log n) - 1)
$
so 
$(1+1/(n \log n))^n - 1
< \log n/((\log n) - 1)-1
= 1/((\log n) - 1)
$.
Putting $a$ for $\log n$, this, combined with
the regular Bernoulli's inequality,
shows that, if $a > 1$ then
$1/a < (1+1/(an))^n-1 < 1/(a-1)$.
A: It's not true.  In fact $$\left(1 + \frac{1}{n \log n}\right)^n = \exp\left(\frac{1}{\log n}\right) + O\left(\frac{1}{n \log n}\right) = 1 + \frac{1}{\log n} + O\left(\frac{1}{\log(n)^2}\right)$$
