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I need to find $\lim_{n \to \infty}$ $(1 + \frac{3}{n^2})^{n^2}$ and I've been given the following:

$\lim_{n \to \infty}$ $n^{1/n}$ = 1, $\lim_{n \to \infty}$ $a^{1/n}$ = 1 and $\lim_{n \to \infty}$ $(1 + \frac{1}{n})^{n}$ = e.

My first thoughts were to use the 3rd limit so $(1 + \frac{3}{n^2})^{n^2}$ <= 3e$^{n}$ and then using the squeeze theorem to show as n tends to infinity the sequence is null, but I think I'm missing something out.

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Hint:Let $N=\frac{n^2}{3}$,then try to use $\lim(1 + \frac{1}{n})^{n}=e$ –  Joe Apr 13 '12 at 12:46
    
@Gingerjin so ${e^(n^{2})/3} $ and as n tends to infinity e = 1? –  franky Apr 13 '12 at 13:01

1 Answer 1

up vote 1 down vote accepted

Hint: $$ \Bigl(1+\textstyle{3\over n^2}\Bigr)^{n^2}=\Bigl(\Bigl(1+{1\over n^3/3}\Bigr)^{n^2/3}\Bigr)^3 $$ Note that $n^2/3\rightarrow\infty$ as $n\rightarrow\infty$.


You might also need to show that, for $x$ a real variable $$\tag{1} \lim_{x\rightarrow\infty} (1+\textstyle{1\over x})^x =e. $$ One way to show this is the following: for $x>1$, $$\textstyle \bigl( 1+{1\over x}\bigr)^x \le \bigl(1+{1\over \lfloor x\rfloor} \bigr)^{\lceil x\rceil}= \bigl(1+{1\over \lfloor x\rfloor} \bigr)^{\lfloor x\rfloor +1}= \bigl(1+{1\over \lfloor x\rfloor} \bigr)^{\lfloor x\rfloor } \bigl(1+{1\over \lfloor x\rfloor} \bigr)^{1} $$ and $$\textstyle \bigl( 1+{1\over x}\bigr)^x \ge \bigl(1+{1\over \lceil x\rceil} \bigr)^{\lfloor x\rfloor} =\bigl(1+{1\over \lceil x\rceil} \bigr)^{\lceil x\rceil -1} =\bigl(1+{1\over \lceil x\rceil} \bigr)^{\lceil x\rceil } \bigl(1+{1\over \lceil x\rceil} \bigr)^{-1}. $$ Apply the Squeeze Theorem to show that $(1)$ holds.

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this eventually makes the solution e^3 right? –  franky Apr 13 '12 at 13:56
    
@franky Yes, indeed. –  David Mitra Apr 13 '12 at 13:57

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