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Let $\{a_n\}$ be a sequence such that $$\lim_{n\to \infty}\left|a_n+3\left(\frac{n-2}{n}\right)^n\right|^\frac{1}{n}=\frac{3}{5}$$

Then calculate $\lim_{n\to \infty}a_n$.

First I tried to take logarithm and got $\lim_{n\to \infty}\frac{1}{n}\ln\left|a_n+3\left(\frac{n-2}{n}\right)^n\right|=\ln\frac{3}{5}$, then I thought about L Hospital but that did not work.

I am unable to dig it further. Can somebody give me a hint or push towards the solution? Thanks.

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  • $\begingroup$ I think limit will be $\infty$ because second term, other than $a_n$ has the limit equal to $\frac{3}{e^2}$. So, if this limit exist, then $\lim_{n\to\infty} a_n=\infty$ $\endgroup$ – Vineet Mangal Jul 7 '16 at 3:10
  • $\begingroup$ Well, $\left(1-\tfrac{2}{n}\right)^n\to e^{-2}$ as $n\to\infty$, so you should use that. $\endgroup$ – John Wayland Bales Jul 7 '16 at 3:10
  • $\begingroup$ you could just set $a_{n}=-3\Big(\frac{n-2}{n}\Big)^{n}+\Big(\frac{3}{5}\Big)^{n}$ and then you can compute the limit quite easily. Question would be, is it unique? $\endgroup$ – Alex Jul 7 '16 at 3:11
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From the given, we can deduce (by the $n$th root test) that the series:

$$\sum \left( a_n + 3\left( \frac{n - 2}n \right)^n \right)$$

is convergent. Therefore,

$$\lim \left( a_n + 3\left( \frac{n - 2}n \right)^n \right) = 0$$

So:

$$\lim a_n = - 3\lim\left( \frac{n - 2}n \right)^n = -3 e^{-2}$$

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