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.

  • $\begingroup$ 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

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,

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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.