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I've got this sequence:

$\quad \displaystyle A_n = \frac{n^3 + n!}{2^n + 3^n}$

And I need to find $\lim_{n\to\infty}A_n$. I've tried using the ratio test and the root test but in this particular case they only seem to make things harder, I think because of the denominator $2^n + 3^n$, which doesn't let me get rid of anything after applying the tests. So any hint about how to tackle this limit will be appreciated.

Thanks in advance!

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3 Answers 3

up vote 3 down vote accepted

For these types of limits (a "rational" form with the limit taken at $\infty$), it usually proves fruitful to divide every term by the highest order term in the denominator. We have $$\def\ts{\displaystyle} A_n={n^3+n!\over 2^n+3^n}={ \ts{n^3\over 3^n}+{n!\over 3^n}\over\ts{2^n\over 3^n} +{3^n\over 3^n}}= {\ts\color{maroon}{n^3\over 3^n}+\color{darkblue}{n!\over 3^n}\over \color{darkgreen}{(2/3)^n}+1}. $$ Now examine each term:

  • $\color{maroon}{\ts\lim\limits_{n\rightarrow\infty}{n^3\over 3^n}=}\ \ \ $ ?
  • $\color{darkgreen}{\ts\lim\limits_{n\rightarrow\infty}{(2/3)^n}=}\ \ \ $ ?
  • $\color{darkblue}{\ts\lim\limits_{n\rightarrow\infty}{n!\over 3^n}=}\ \ \ $ ?

Once you've computed the above limits, you should be able to evaluate the original limit.

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I'll take this tip into accout, thanks @DavidMitra ! –  Lucas Jan 30 '13 at 17:39

if you quit the $n^3$ the number will be smaller, and if you changer the $2^n$ by a $3^n$ down in the fraction, the number will smaller too, so $$A_n > \frac{n!}{3^{n+1}}$$ for a sufficiently large $n$. If you can get the limit of the right hand sequence, you're done.

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Clearly there are lots of ways to solve an exercise, thank you for your time @MyUserIsThis. –  Lucas Jan 30 '13 at 17:46

Hint:

$$A_n \geq \frac{n!}{2\cdot 3^n}$$

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