ISI2015-MMA-10

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The value of the infinite product

$$P=\frac{7}{9} \times \frac{26}{28} \times \frac{63}{65} \times \cdots \times \frac{n^3-1}{n^3+1} \times \cdots \text{ is }$$

1. $1$
2. $2/3$
3. $7/3$
4. none of the above
in Calculus
retagged
0
$\frac{2}{3}$ ??

Here, \begin{align}P =\lim_{n\to \infty}\prod_{r=2}^{n} \frac{r^3-1}{r^3+1}&=\lim_{n\to \infty}\prod_{r=2}^{n} \left(\frac{r-1}{r+1}\right)\left(\frac{r^2+r+1}{r^2-r+1}\right)\\&=\lim_{n\to \infty}\left(\prod_{r=2}^{n} \frac{r-1}{r+1}\right) \left( \prod_{r=2}^{n} \frac{r^2+r+1}{r^2-r+1}\right)......\tag{i}\end{align}

Now

\begin{align} \prod_{r=2}^{n} \frac{r-1}{r+1}&=\frac{1}{3}\times\frac{2}{4}\times\frac{3}{5}\times\frac{4}{6}\times\frac{5}{7}\times\frac{6}{8}\times\cdots\times\frac{n-2}{n}\times\frac{n-1}{n+1}\\&=\frac{1\times2}{n(n+1)}; [\scriptsize \text{All numbers, except the first two numerators and the last two denominators,}\\&~~~~~~~~~~~~~~~~~~~~~~\scriptsize\text{ are crossed out by each other.}]\\&=\frac{2}{n(n+1)} \end{align}

Then

\begin{align} &~~~~\prod_{r=2}^{n} \frac{r^2+r+1}{r^2-r+1}\\&=\prod_{r=2}^{n} \frac{r(r+1)+1}{r(r-1)+1}\\&=\scriptsize\frac{(2\times3+1)}{(2\times1+1)}\times\frac{(3\times4+1)}{(3\times2+1)}\times\frac{(4\times5+1)}{(4\times3+1)}\times\cdots\times\frac{(n-1)n+1}{(n-1)(n-2)+1}\times\frac{n(n+1)+1}{n(n-1)+1}\\&=\frac{n(n+1)+1}{2\times1+1}; [\scriptsize \text{All numbers, except the first denominator and the last numerator,}\\&~~~~~~~~~~~~~~~~~~~~~~~~~~~~\scriptsize\text{ are crossed out by each other.}]\\&=\frac{n(n+1)+1}{3} \end{align}

Now putting these values to no$\mathrm{(i)}$, we get

\begin{align}P &=\lim_{n\to \infty}\left(\frac{2}{n(n+1)}\right) \left( \frac{n(n+1)+1}{3}\right)\\&=\frac{2}{3}\lim_{n\to \infty} \left( 1+\frac{1}{n(n+1)} \right)\\&=\frac{2}{3}; ~~[\because n\to \infty \Rightarrow \frac{1}{n}\to 0 \Rightarrow \frac{1}{n(n+1)}\to 0] \end{align}

So the correct answer is B.

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Answer: $\mathbf B$

Given:$$P = \frac{7}{9} \times \frac{26}{28} \times \frac{63}{65}\times\dots\times\frac{n^3-1}{n^3+1}\times\dots$$

This can be written as,

$$\implies P_n \require {cancel} \require {circle}= \Big (\frac{1}{\color {red} 3} \times \frac{\cancel 7}{\cancel3}\Big)\times \Big(\frac{\color {red}2}{\cancel 4}\times\frac{\cancel{13}}{\cancel 7})\times\Big(\frac{\cancel 3}{\cancel 5}\times\frac{\cancel {21}}{\cancel {13}}\Big)\times\Big(\frac{\cancel4}{\cancel 7}\dots\times\Big(\frac{n-1}{n+1}\times \frac{n^2+n+1}{n^2-n+1}\Big)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\scriptsize [\text{So, everything will cancel out except 2 and 3}]$$

$$\implies P_n = \frac{2}{3} \times \frac{n^2+n+1}{n(n+1)} =\frac{2}{3}\times\Big(1 + \frac{1}{n(n+1)}\Big)$$

Now, as $$n \to \infty, \;P_n \to \frac{2}{3}$$

$\therefore$ the infinite product, $P$, converges to $\frac{2}{3}$; $\implies P_\infty = \frac{2}{3}$

$\therefore \; \mathbf B$ is the correct answer.

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