TITLE: Finding the limit of a sequence $a_n$
QUESTION [2 upvotes]: 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.

REPLY [0 votes]: $$a_n = n^{3}-\left(n^{3}+n-1\right)\left(1-\frac{1}{n^{3}}\right)^{n-1}$$
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 $n \sim 49$.
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