What is the value of lim$_{n\to \infty} a_n$ if $\lim_{n \to \infty}\left|a_n+ 3\left(\frac{n-2}{n}\right)^n \right|^{\frac{1}{n}}=\frac{3}{5}$? Let $\{a_n\}$ be a sequence of real numbers such that $$\lim_{n \to \infty}\left|a_n+ 3\left(\frac{n-2}{n}\right)^n \right|^{\frac{1}{n}}=\frac{3}{5}.$$
What is the value of $\lim_{n\to \infty} a_n$?
 A: We have $$\lim_{n \to \infty}|a_n+ 3((n-2)/n)^n|^{1/n}=3/5 <1.$$ So by the root test, the series $\sum|(a_n+ 3((n-2)/n)^n|$ converges, therefore the general term converges to $0$.
Note that:
$$((n-2)/n)^n = (1-2/n)^n \to e^{-2}$$
This shows that $a_n \to -3e^{-2}$.
A: If $\displaystyle \lim_{n \to \infty}\left|a_n+ 3\left(1-\frac2n\right)^n\right|^{1/n}=3/5$, then for all $\epsilon>0$, there exists a number $N(\epsilon)$ such that 
$$\frac35-\epsilon\le\left|a_n+ 3\left(1-\frac2n\right)^n\right|^{1/n}<\frac35+\epsilon \tag 1$$
whenever $n>N(\epsilon)$.  The inequality in $(1)$ is equivalent to the inequality
$$\left(\frac35-\epsilon\right)^n\le \left|a_n+ 3\left(1-\frac2n\right)^n\right|\le \left(\frac35+\epsilon\right)^n \tag 2$$
for $n>N(\epsilon)$.
Since $(2)$ is true for all $\epsilon$, it is true for $\epsilon\le r<\frac25$.  Letting $n\to \infty$ in $(2)$ for such $\epsilon$, we see that 
$$\lim_{n\to \infty}\left|a_n+ 3\left(1-\frac2n\right)^n\right|=0.$$
Since the absolute value function is continuous, we must have
$$\lim_{n\to \infty}\left(a_n+ 3\left(1-\frac2n\right)^n\right)=0,$$
which implies
$$\lim_{n\to \infty}a_n=-\lim_{n\to \infty}\left(3\left(1-\frac2n\right)^n\right)=-3e^{-2}.$$
And we are done!
A: We have
$$\lim_{n\to +\infty}3\left(\frac{n-2}{n}\right)^n=3e^{-2}$$
$$\implies \lim_{n \to\infty}a_n=-3e^{-2}$$
since $$\lim_{n\to+\infty}\left(\frac{3}{5}\right)^n=0.$$
