# 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 $\lim_{n\to \infty} a_n$?

• I edited your question. Some LaTeX tips: you can use \{ \} inside math mode (i.e., inside the $signs) to make braces. You can use \lim to make the limit symbol. Commented Nov 27, 2016 at 16:02 • – user9464 Commented Nov 27, 2016 at 16:24 ## 3 Answers 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$a_n \to -3e^{-2}$. • @Dr.MV when a series converges, its general term goes to$0$– user384138 Commented Nov 27, 2016 at 16:24 • Yes, I know. But we need not use results from series analysis when analyzing a sequence. Commented Nov 27, 2016 at 16:27 • @Dr.MV but then we have to basically prove that thing (which you did in your answer) – user384138 Commented Nov 27, 2016 at 16:29 • Your proof is fine. It's quite possible that the OP and others who are in the middle of studying sequences have not yet studied series. They would not have, therefore, the requisite tools to apply this approach. That is the reason that I took a "back to basics" approach. Commented Nov 27, 2016 at 16:32 • Good approach+1. Commented Nov 28, 2016 at 16:37 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!

• You are basically proving the known result I used in my answer (+1)
– user384138
Commented Nov 27, 2016 at 16:27
• @OpenBall Thank you for the "up vote!" Much appreciative. Commented Nov 27, 2016 at 16:28

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 $$\lim_{n\to+\infty}\left(\frac{3}{5}\right)^n=0.$$

• that is very clean.............+1 Commented Dec 11, 2016 at 12:09
• @Bhaskara-III Thanks a lot for your comment. Commented Dec 11, 2016 at 12:11