# More highschool math $\frac{3^n(n^3+3n^2+3n+1)}{3^{n+1}\cdot n^3} = \frac{n^3+3n^2+3n+1}{3n^3} \to \frac{1}{3}$

So the question I am trying to work through is:

Test the series
$$\frac{1}{3}+\frac{2^3}{3^2}+\frac{3^3}{3^3}+\frac{4^3}{3^4}+\frac{5^3}{3^5}+\cdot\cdot\cdot$$ for convergence.

The solution (using D'Alembert's ratio test) is:
$$u_n=\frac{n^3}{3^n}\;,$$

so

\begin{align*} \frac{|u_{n+1}|}{|u_n|} &=\frac{(n+1)^3}{3^{n+1}}\cdot \frac{3^n}{n^3}\\ &=\frac{3^n(n+1)^3}{3^{n+1}\cdot n^3}\\ &=\frac{3^n(n+1)^3}{3^{n+1}\cdot n^3}\\ &=\frac{3^n(n^3+3n^2+3n+1)}{3^{n+1}\cdot n^3}\;. \end{align*}

How do we get from there to...

$$=\frac{n^3+3n^2+3n+1}{3n^3}$$

What happens with $3^n$ in the numerator and power of $n+1$ in the denominator? How do they cancel out?

Also, in the very next step that all goes to being equal to

$$\lim\limits_{n\rightarrow\infty}\frac{|u_{n+1}|}{|u_n|}=\frac{1}{3}<1\;,$$

which means the series is convergent.

But how do we get to $\dfrac{1}{3}$?

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only in the limit $n \rightarrow \infty$ your expression is equal to $1/3$ to see this simple divide the numerator and denominator by $n^{3}$ – Jose Garcia Apr 29 '12 at 21:22
3^n/3^(n+1)=1/3 – dot dot Apr 29 '12 at 21:24
$$\frac{3^n(n^3+3n^2+3n+1)}{3^{n+1}n^3}=\frac{3^n}{3^{n+1}}\cdot\frac{n^3+3n^2+3‌​n+1}{n^3}=\frac13\left(1+\frac3n+\frac3{n^2}+\frac1{n^3}\right)\;.$$ Now take the limit as $n\to\infty$, as @Jose said. – Brian M. Scott Apr 29 '12 at 21:26
@Gineer Note that is not technically correct to write $$\frac{n^3+3n^2+3n+1}{3n^3} = \frac{1}{3}$$ What you might want to suggest in the notation is $$\frac{n^3+3n^2+3n+1}{3n^3} \to \frac{1}{3}$$ or $$\frac{n^3+3n^2+3n+1}{3n^3} \mathop = \limits^{n \to \infty } \frac{1}{3}$$ – Pedro Tamaroff Apr 29 '12 at 22:01

1. You have a factor of $3^n$ in the numerator, and a factor of $3^{n+1}$ in the denominator. So $$\frac{3^n(\text{stuff})}{3^{n+1}(\text{other stuff})} = \frac{3^n(\text{stuff})}{3\times 3^{n}\text{(other stuff)}} = \frac{\text{stuff}}{3(\text{other stuff})}.$$ Since $3^{n+1}=3\times 3^n$.

2. Dividing numerator and denominator by $n^3$, we have \begin{align*} \lim_{n\to\infty}\frac{n^3+3n^2+3n+1}{3n^3} &= \lim_{n\to\infty}\frac{\frac{1}{n^3}(n^3+3n^2+3n+1)}{\frac{1}{n^3}(3n^3)}\\ &= \lim_{n\to\infty}\frac{1 + \frac{3}{n} + \frac{3}{n^2}+\frac{1}{n^3}}{3}\\ &= \frac{\lim\limits_{n\to\infty}(1 + \frac{3}{n}+\frac{3}{n^2}+\frac{1}{n^3})}{\lim\limits_{n\to\infty}3}\\ &= \frac{1 + 0 + 0 + 0}{3} = \frac{1}{3}. \end{align*}

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Here is one way to simplify the limit and arrive at the answer. Hopefully, it will let you see how terms cancel out.

\begin{align*} \left|\frac{u_{n+1}}{u_{n}}\right| &= \frac{(n+1)^3}{3^{n+1}}\cdot \frac{3^n}{n^3} \\ &= \frac{(n+1)^3}{n^3}\cdot \frac{3^n}{3^{n+1}} \\ &= \left(\frac{n+1}{n}\right)^3 \cdot \frac{1}{3} \\ &= \left(1 + \frac{1}{n}\right)^3 \cdot \frac{1}{3} \\ \Rightarrow \lim_{n \rightarrow \infty} \left|\frac{u_{n+1}}{u_{n}}\right| &= 1^3 \cdot \frac{1}{3} = \frac{1}{3} \end{align*}

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What happens with the $3^n$ in the numerator and the $3^{n+1}$ in the denominator?

Recall the following laws of exponents: $$a^{b}a^c = a^{b+c}, \quad \text{ and } \quad \frac{a^b}{a^c} = a^{b-c} = \frac{1}{a^{c-b}},$$ for any $a>0$, and any real numbers $b,c$. In particular, if $a=3$, $b=n$, and $c=n+1$, then: $$\frac{3^n}{3^{n+1}}=\frac{1}{3^{(n+1)-n}} = \frac{1}{3^1}=\frac{1}{3}.$$

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