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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:


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

How do we get from there to...


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


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
up vote 5 down vote accepted
  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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