# Find the Limit?

## ${\lim}_{n \rightarrow \infty} {P}_{n}$ where P_n=(2^3-1)/(2^3+1)*(3^3-1)/(3^3+1)*.....*(n^3-1)/(n^3+1); n=2,3,...

Apr 7, 2018

${\lim}_{n \to \infty} {P}_{n} = \frac{2}{3}$

#### Explanation:

We see that

${P}_{n} = {\prod}_{k = 2}^{n} \frac{{k}^{3} - 1}{{k}^{3} + 1} = {\prod}_{k = 2}^{n} \frac{{k}^{3} - {1}^{3}}{{k}^{3} + {1}^{3}}$

Both the numerator and denominator are a sum or difference of cubes, for which we have a formula:

${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$
${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

Thus, we have

${P}_{n} = {\prod}_{k = 2}^{n} \frac{\left(k - 1\right) \left({k}^{2} + k + 1\right)}{\left(k + 1\right) \left({k}^{2} - k + 1\right)}$

${P}_{n} = \frac{\left(2 - 1\right) \left({2}^{2} + 2 + 1\right)}{\left(2 + 1\right) \left({2}^{2} - 2 + 1\right)} \cdot \frac{\left(3 - 1\right) \left({3}^{2} + 3 + 1\right)}{\left(3 + 1\right) \left({3}^{2} - 3 + 1\right)} \ldots \frac{\left(n - 2\right) \left({n}^{2} - n + 1\right)}{\left(n\right) \left({n}^{2} - 3 n + 3\right)} \frac{\left(n - 1\right) \left({n}^{2} + n + 1\right)}{\left(n + 1\right) \left({n}^{2} - n + 1\right)}$

We can group ${P}_{n}$ into the following products:

${P}_{n} = \textcolor{red}{{\prod}_{k = 2}^{n} \frac{k - 1}{k + 1}} \cdot \left(\frac{{2}^{2} + 2 + 1}{{2}^{2} - 2 + 1}\right) \left(\frac{{3}^{2} + 3 + 1}{{3}^{2} - 3 + 1}\right) \ldots \left(\frac{{n}^{2} - n + 1}{{n}^{2} - 3 n + 3}\right) \left(\frac{{n}^{2} + n + 1}{{n}^{2} - n + 1}\right)$

Let the product in $\textcolor{red}{\text{red}}$ be ${p}_{n}$.

Notice how some things cancel each other:

${P}_{n} = \textcolor{red}{{p}_{n}} \cdot \left(\frac{\cancel{{2}^{2} + 2 + 1}}{{2}^{2} - 2 + 1}\right) \left(\frac{\cancel{{3}^{2} + 3 + 1}}{\cancel{{3}^{2} - 3 + 1}}\right) \ldots \left(\frac{\cancel{{n}^{2} - n + 1}}{\cancel{{n}^{2} - 3 n + 3}}\right) \left(\frac{{n}^{2} + n + 1}{\cancel{{n}^{2} - n + 1}}\right)$

It might seem confusing, but the numerator of the $k$-th term is being cancelled by the denominator of the $k + 1$-th term.

$\therefore$

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

Let's write out ${p}_{n}$:

${p}_{n} = \frac{2 - 1}{2 + 1} \cdot \frac{3 - 1}{3 + 1} \ldots \frac{n - 2}{n} \cdot \frac{n - 1}{n + 1}$

${p}_{n} = \frac{1 \cdot 2 \cdot 3 \cdot \ldots \cdot \left(n - 2\right) \left(n - 1\right)}{3 \cdot 4 \cdot 5 \cdot \ldots \cdot n \left(n + 1\right)}$

Again, lots of things cancel out.

${p}_{n} = \frac{1 \cdot 2 \cdot \cancel{3 \cdot 4 \cdot 5 \cdot \ldots \cdot \left(n - 2\right) \left(n - 1\right)}}{\cancel{3 \cdot 4 \cdot 5 \cdot \ldots \cdot \left(n - 2\right) \left(n - 1\right)} n \left(n + 1\right)} = \frac{2}{n \left(n + 1\right)}$

Therefore, we have:

${P}_{n} = \textcolor{red}{\frac{2}{n \left(n + 1\right)}} \cdot \left[\frac{n \left(n + 1\right)}{3} + \frac{1}{3}\right]$

${P}_{n} = \frac{2}{3} \left[1 + \frac{1}{n \left(n + 1\right)}\right]$

Now, let's take the $\textcolor{red}{\text{limit}}$ as $n \to \infty$ of both sides.

${\lim}_{n \to \infty} {P}_{n} = {\lim}_{n \to \infty} \frac{2}{3} \left[1 + \frac{1}{n \left(n + 1\right)}\right]$

As $n$ grows boundless, $\frac{1}{n \left(n + 1\right)}$ becomes arbitrarily small.

In other words, it approaches $0$.

$\implies {\lim}_{n \to \infty} {P}_{n} = \frac{2}{3} \left[1 + 0\right] = \frac{2}{3}$

$\textcolor{red}{{\lim}_{n \to \infty} {P}_{n} = \frac{2}{3}}$