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Define $a_n$ as:

$$ a_n = \left(\left(1+\frac{1}{n^2}\right)^{n^2}\left(1-\frac{1}{n}\right)^n\left(1+\frac{1}{n}\right)\right)^n $$

Now I want to calculate $\lim_{n \to \infty} a_n$. So, real question is about other limit.

$$ \lim_{n\to\infty}\left(\left(1+\frac{1}{n^2}\right)^{n^2} \left(1-\frac{1}{n}\right)^n\right)^n $$

As $\lim_{n\to\infty}\left(1+\frac{1}{n^2}\right)^{n^2} \left(1-\frac{1}{n}\right)^n = 1$, we cannot conclude limit in easy way. ($1^{\infty}$ case.) I don't know, what I should do now.

I tried:

$$ \ln \left(\left(1+\frac{1}{n^2}\right)^{n^2} \left(1-\frac{1}{n}\right)^n\right) = n \left(n^2\ln(1+\frac{1}{n^2}) + n \ln(1-\frac{1}{n})\right) $$

So, now problematic is limit:

$$ \lim_{n \to \infty} n \left(n^2\ln(1+\frac{1}{n^2}) + n \ln(1-\frac{1}{n})\right) $$

It's not better. How should I calculate this limit?

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  • $\begingroup$ Hint: $\lim_{n\to\infty}(1+\frac{1}{n^2})^{n^2}=e$ and $\lim_{n\to\infty}(1-\frac{1}{n})^{n}= \frac 1e$ $\endgroup$
    – user275377
    Apr 7 '16 at 21:36
  • $\begingroup$ True, but I don't see, how it should help. $\endgroup$
    – Tacet
    Apr 7 '16 at 21:38
  • $\begingroup$ Write as $\lim_{n \to \infty} \frac {\ln{((1+ \frac 1{n^2})^{n^2}(1- \frac 1n)^n)}}{n^{-1}}$ then you have a "$\frac 00$" case, apply l'hopitals rule $\endgroup$
    – user275377
    Apr 7 '16 at 22:05
  • $\begingroup$ @Alex I'm not sure if it can work. It's very unpleasant derivative. If you see a way to calculate it by hand that way. Please, submit your answer. $\endgroup$
    – Tacet
    Apr 7 '16 at 22:11
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Let $f_n$ be the sequence given by $f_n=n \left(n^2\log\left(1+\frac{1}{n^2}\right) + n \log\left(1-\frac{1}{n}\right)\right)$. We can evaluate the limit of $f_n$ by expanding the logarithm functions embedded in the limit of interest as

$$\begin{align} \lim_{n\to \infty}f_n&=\lim_{n \to \infty} n \left(n^2\log\left(1+\frac{1}{n^2}\right) + n \log\left(1-\frac{1}{n}\right)\right)\\\\ &=\lim_{n \to \infty} n\left(n^2\left(\frac{1}{n^2}+O\left(\frac{1}{n^4}\right)\right)+n\left(-\frac{1}{n}-\frac{1}{2n^2}+O\left(\frac{1}{n^3}\right)\right)\right)\\\\ &=-\frac12 \end{align}$$

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  • $\begingroup$ I just saw it. However, thank you for your answer. $\endgroup$
    – Tacet
    Apr 7 '16 at 21:45
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    $\begingroup$ You're quite welcome. My pleasure! $\endgroup$
    – Mark Viola
    Apr 7 '16 at 21:46

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