# Finding limit of the sequence $A_n = \frac{n^3 + n!}{2^n + 3^n}$

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.

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.

• 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.

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

Hint:

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