Finding the limit of a sequence $a_n$ I have the following sequence $$ a_n = n^{3}-\left(n^{3}+n-1\right)\left(1-\frac{1}{n^{3}}\right)^{n-1} $$ I want to find $\lim_{n\to\infty}a_n$, I know it is 0, but I can't seem to find a way to show it. It seems like $\left(n^{3}+n-1\right)\left(1-\frac{1}{n^{3}}\right)^{n-1}$ is an (good) approximation of $x^3$, but it isn't a taylor series or anything like it I know of. Since it is ( / it looks like) an appromixation, once I try to apply the sandwich rule, I get large numbers, which don't help me. It also seems that Cauchy criteria doesn't make it easier.
 A: HINT:
From Bernoulli's Inequality,
$$\frac1{1+\frac{n-1}{n^3}}\ge \left(1-\frac1{n^3}\right)^{n-1}\ge 1-\frac{n-1}{n^3}$$
Use these inequalities and apply the squeeze theorem.
Can you finish now?
A: We can use that $\log(1+x)=x+O(x^2)$ and $e^x=1+x+O(x^2)$ as $x \to 0$ to obtain
$$\left(1-\frac{1}{n^{3}}\right)^{n-1}=e^{(n-1)\log\left(1-\frac{1}{n^{3}}\right)}=e^{-\frac{n-1}{n^3}+O\left(\frac1{n^5}\right)}=1-\frac{n-1}{n^3}+O\left(\frac1{n^5}\right)$$
and therefore
$$n^{3}-\left(n^{3}+n-1\right)\left(1-\frac{1}{n^{3}}\right)^{n-1}=n^{3}-\left(n^{3}+n-1\right)\left(1-\frac{n-1}{n^3}+O\left(\frac1{n^5}\right)\right)$$
A: Hint: Rewrite your sequence as $$a_{n} = n^{3}- n^{3}\left(\frac{n^{3}+n-1}{n^{3}-1}\right)\left(\frac{n^{3}-1}{n^{3}}\right)^{n}$$
A: $$a_n = n^{3}-\left(n^{3}+n-1\right)\left(1-\frac{1}{n^{3}}\right)^{n-1}$$
We need to focus on the last term
$$b_n=\left(1-\frac{1}{n^{3}}\right)^{n-1}\implies \log(b_n)=(n-1)\log\left(1-\frac{1}{n^{3}}\right)$$ By Taylor
$$\log\left(1-\frac{1}{n^{3}}\right)=-\frac{1}{n^3}-\frac{1}{2 n^6}+O\left(\frac{1}{n^{9}}\right)$$
$$\log(b_n)=-\frac{1}{n^2}+\frac{1}{n^3}-\frac{1}{2 n^5}+\frac{1}{2
   n^6}+O\left(\frac{1}{n^{8}}\right)$$
$$b_n=e^{\log(b_n)}=1-\frac{1}{n^2}+\frac{1}{n^3}+\frac{1}{2 n^4}-\frac{3}{2
   n^5}+\frac{5}{6
   n^6}+\frac{1}{n^7}+O\left(\frac{1}{n^8}\right)$$
All of that makes
$$a_n=\frac{1}{2 n}-\frac{1}{2 n^2}-\frac{1}{3
   n^3}+\frac{1}{n^4}+O\left(\frac{1}{n^5}\right)$$ which shows the limit but also that the limit is approched from above.
Moreover, it allows a fast way to find the approximate value of $n$ for a given value of $a_n$. Suppose $a_n=0.01$ so the first term of the expansion gives $n \sim 50$; using the first and second terms gives $n \sim 49$.
