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?


You cannot write things like

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

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), $\lim_{n\to +\infty}a_n=-3e^{-2}$.

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  • $\begingroup$ Okay thanks. But can you throw more light on why right hand side should not depend on $n$ being limit over $n$? $\endgroup$ – Error 404 Jan 24 '16 at 9:41
  • $\begingroup$ The limit should be fixed number since a sequence of real numbers cannot converge to two different limits. $\endgroup$ – Davide Giraudo Jan 24 '16 at 9:42
  • $\begingroup$ Okay. Thank you very much. $\endgroup$ – Error 404 Jan 24 '16 at 9:43

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