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How would you prove the following? $$\lim_{n \rightarrow \infty} (1+c/n^2 - 2/n)^n = e^{-2}$$

where $c\geq 1$ is a constant.

I can find proofs online of $\lim_{n \rightarrow \infty} (1-2/n)^n = e^{-2}$ but how do you make the above rigorous?

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Hint: $(1 + \frac{1}{n^2} - \frac{2}{n}) = (1-\frac{1}{n})^2$. –  froggie Dec 13 '12 at 14:31
    
Edited question slightly (but thanks). –  Sprong Dec 13 '12 at 14:35

2 Answers 2

up vote 4 down vote accepted

Note that

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

ADD

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

so that $$y_n=\log x_n=n \log \left(1- \frac 2 n+\frac c {n^2}\right)$$

Then

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

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

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

Now $a_n\to 0$; so that

$$\left(\frac c {n}- 2\right)\frac{\log \left(1+a_n\right)}{a_n}\to (0-2)\cdot 1 =-2$$

This means $x_n\to e^{-2}$

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Edited question to make it clearer that I am interested in not just $c=1$. –  Sprong Dec 13 '12 at 14:33
    
@Sprong Thanks. You need to use that $\log(1+x)/x$ tends to $1$ which also needs a proof I suppose. –  Sprong Dec 13 '12 at 15:09
    
I am using that. You can find various proofs of that, or give one yourself. –  Pedro Tamaroff Dec 13 '12 at 15:18

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

$$= \left(\lim _{n\to\infty}\left(1+\frac{c-2n}{n^2}\right)^{\frac{n^2}{c-2n}}\right)^{\lim _{n\to\infty}\left(\frac{c-2n}n\right)}$$

$$= \left(\lim _{m\to\infty}\left(1+\frac1m\right)^m\right)^{\lim _{n\to\infty}\left(\frac{c-2n}n\right)}$$ where $m=\frac{n^2}{c-2n},n\to \infty\implies m\to\infty$

$$=e^{-2}$$ for any finite value of $c$

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Can you explain this a little more please? –  Sprong Dec 13 '12 at 14:40
    
@Sprong, please find the edited answer. –  lab bhattacharjee Dec 13 '12 at 14:44
    
I am a little worried about the part where you put a limit also in the exponent. How do we know that is valid? –  Sprong Dec 13 '12 at 14:45
    
@Sprong, please use logarithm if you are not comfortable with exponent. –  lab bhattacharjee Dec 13 '12 at 14:46

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