Limit of $\{a_n\}$ where $\lim_{n \to \infty}\left|a_n+3\left(1-\frac 2n\right)^n\right|^{\frac 1n}=\frac 35$

Taking $n$-th power both sides,

we get $\lim_{n \to \infty}\left|a_n+3\left(1-\frac 2n\right)^n\right|=\left(\frac 35\right)^n$

Let $\lim_{n \to \infty}a_n=t$.

Then applying the limits we get $\left|t+\frac 3{e^2}\right|=0$, which gives $t=- \frac 3{e^2}.$

Is this approach fine?

$$\lim_{n \to \infty}\left|a_n+3\left(1-\frac 2n\right)^n\right|=\left(\frac 35\right)^n$$
because the right hand side should not depend on $n$ (this is a limit over $n$). However, you can make this idea rigorous using the fact that $$\left(\left|a_n+3\left(1-\frac 2n\right)^n\right|\right)^{1/n}\leqslant \frac 45$$ for $n$ large enough (why?). This gives that $$\lim_{n\to +\infty}a_n+3\left(1-\frac 2n\right)^n=0,$$ hence (as you found), $\lim_{n\to +\infty}a_n=-3e^{-2}$.
• Okay thanks. But can you throw more light on why right hand side should not depend on $n$ being limit over $n$? – Error 404 Jan 24 '16 at 9:41