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Find if the sequence $a_n$ converges, $a_n=\frac{n!}{10^{6n}}=\frac{1}{\frac{10^6n}{n!}}$. Now if we take the limit of numerator and denominator as $n \to \infty$, we get $\frac{1}{0}$($\lim\limits_{n\to\infty} \frac{x^n}{n!}=0 $ for any $x$), but division by zero is not defined, so we are stuck, but the answer is $\infty$. Where am I going wrong?

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Fact: If $x_n\gt0$ and $x_n\to0$, then $1/x_n\to+\infty$. Can you show this fact and apply it to your setting? –  Did Apr 2 '12 at 6:08
    
Look at the argument which establishes the limit you know (the one which tends to zero) and see if you can adapt it directly to the new case. Alternatively, what does that limit mean - you can make a certain expression as close to zero as you choose: how does that help you with the reciprocal? –  Mark Bennet Apr 2 '12 at 6:08

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up vote 4 down vote accepted

Let $$b_n=\frac1{a_n}=\frac{10^{6n}}{n!}\;.$$ It appears from what you say in the question that you know that $$\lim_{n\to\infty}b_n=0\;.$$ It’s also clear that $b_n>0$ for all $n$. Thus, as $n\to\infty$, we have not just that $b_n\to 0$, but more specifically that $b_n\to 0^+$. But then clearly $$a_n=\frac1{b_n}\to+\infty\;.$$ At no point should you be trying to divide by $0$; that manipulation is clearly illegitimate.

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Hint for an alternative route,

$$\rm \frac{a_{\large n+10^6}}{a_{\large 10^6}}\normalsize =\frac{10^6+1}{10^6}\cdot\frac{10^6+2}{10^6}\cdots\frac{10^6+n}{10^6}>\left(1+\frac{1}{10^6}\right)^n \to \infty.$$

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