Convergence of series, using big oh or little oh notation. Let $p\in \mathbb{R}$ and $a_n=(e-(1+1/n)^n)^p$. For which $p$ will $\sum_{n=1}^{\infty} a_n$ converges?
Because of the "additive look" of $a_n$, I tried to use taylor expansion and big oh, little oh notation to solve the problem but I couldnt solve it, please helps. Below is what i have done so far:
Observe that $\ln (1+x)=x+O(x^2)$ and $1-e^x \sim -x$. Hence
$$(1+1/n)^n=e^{n \ln (1+1/n)}=e^{n(1/n+O(1/n^2))}$$ and 
$$e-(1+1/n)^n=e-e^{1+O(1/n)}=e(1-e^{O(1/n)}) \sim e(-O(1/n))=O(1/n).$$
So 
$$a_n= (e-(1+1/n)^n)^p \sim O(1/n)^p=O(1/n^p)$$
But then I dont know how to move on...

That was quite good, but you need to go deeper if you want to solve by just using asymptotics.
$\displaystyle e^1-\exp\left(n\ln\left(1+\frac{1}{n}\right)\right)=\\e^1-\exp\left(n\left(\frac{1}{n}-\frac{1}{n^2}+o\left(\frac{1}{n^2}\right)\right)\right)=\\e^1\times\left(1-\exp\left(\frac{-1}n+o\left(\frac{1}{n}\right)\right)\right)=\\\displaystyle e^1\times\left(1-\left(1-\frac{1}{n}+o\left(\frac{1}{n}\right)\right)\right)=\\e^1\times \left(\frac{1}{n}+o\left(\frac{1}{n}\right)\right). $
Therefore, $\displaystyle a_n\sim\frac{e^1}{n^p}$, which is more powerful than a mere $a_n=O(1/n^p)$
You should already know that when $a_n\sim b_n$ and $a_n > 0$ then $\sum a_n$ converges iff $\sum b_n$ converges.
Therefore, $\sum a_n$ converges iff $\sum 1/n^p$, that is iff $p>1$.
