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I have this series

$\prod_{k=2}^n\frac{k^2-1}{k^2}$

and I am asked to write an equation describing the partial products. Since I know that this isn't a geometric or an arithmetic sequence, I'm stuck. Can someone help out? Thanks in advance!

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    I rolled it back. Your edit was messed up. Try again.2012-05-29

3 Answers 3

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You can cleverly get a solution by working as follows:

$\prod\limits_{k = 2}^n {\frac{{{k^2} - 1}}{{{k^2}}}} = \prod\limits_{k = 2}^n {\frac{{k - 1}}{k}\frac{{k + 1}}{k}} = \prod\limits_{k = 2}^n {\frac{{k - 1}}{k}} \prod\limits_{k = 2}^n {\frac{{k + 1}}{k}} = \prod\limits_{k = 2}^n {\frac{{{a_{k - 1}}}}{{{a_k}}}} \prod\limits_{k = 2}^n {\frac{{{a_{k + 1}}}}{{{a_k}}}} $

with $a_k=k$

Now use the telescopic rule of products:

$\prod\limits_{k = 2}^n {\frac{{{a_{k - 1}}}}{{{a_k}}}} = \frac{{{a_1}}}{{{a_n}}}$

$\prod\limits_{k = 2}^n {\frac{{{a_{k + 1}}}}{{{a_k}}}} = \frac{{{a_{n + 1}}}}{{{a_2}}}$

this gives

$\prod\limits_{k = 2}^n {\frac{{{k^2} - 1}}{{{k^2}}}} = \frac{1}{2}\frac{{n + 1}}{n}$

so when $n\to \infty$, the answer is

$\prod\limits_{k = 2}^\infty {\frac{{{k^2} - 1}}{{{k^2}}}} = \frac{1}{2}$

I really hope this is what you're asking about.

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Look at the partial sum i.e. $S_n = \sum_{x=2}^{n} \dfrac{x^2-1}{x^2} = \sum_{x=1}^{n} \dfrac{x^2-1}{x^2} = \sum_{x=1}^{n} \left(1 - \dfrac1{x^2} \right) = n - \sum_{x=1}^{n} \dfrac1{x^2}$ If you want to compute $S_n$ for a given $n$, then it essentially boils down to computing $\displaystyle \sum_{x=1}^{n} \dfrac1{x^2}$. There is no explicit formula as a function of $n$ which can be evaluated easily for this sum, however that doesn't stop us from computing the sum approximately. The Euler Maclaurin formula is usually used to approximately evaluate such sums.

From the Euler Maclaurin summation, you get that $\sum_{k=1}^n f(k) = \int_1^{n} f(x)dx - B_1 \left( f(n) + f(1)\right) + \sum_{k=1}^{r} \dfrac{B_{2k}}{(2k)!} \left(f^{(2k-1)}(n) - f^{(2k-1)}(1) \right) + R(r)$ where the remainder $R(r)$ can be bounded as $\lvert R(r)\rvert \leq \dfrac{2 \zeta(2r)}{(2 \pi)^{2r}} \int_1^{n} \left \lvert f^{2r}(x) \right \rvert dx$

EDIT If you want the infinite product, then look at $S_n = \prod_{x=2}^{n} \dfrac{x^2-1}{x^2} = \prod_{x=2}^{n} \dfrac{(x-1)(x+1)}{x^2} = \dfrac{1 \times 3}{2^2} \dfrac{2 \times 4}{3^2}\dfrac{3 \times 5}{4^2}\dfrac{4 \times 6}{5^2} \cdots \dfrac{(n-1) \times (n+1)}{n^2} = \dfrac{1}{2} \dfrac{(n+1)}{n}$ Hence, $\prod_{x=2}^{\infty}\dfrac{x^2-1}{x^2} = \lim_{n \rightarrow \infty} S_n = \dfrac12$

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    Seems the OP now wants an infinite product!2012-05-29
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As per the comments, the question now is to find a formula for $\sum_{n=2}^{11}{x^2-1\over x^2}$ Clearly this is equivalent to finding a formula for $\sum_{n=2}^{11}{1\over x^2}$ or more generally $\sum_{n=2}^Q{1\over x^2}$ There is no closed-form formula for this sum in terms of the familiar functions of intro calculus (powers, logarithms, exponentials, trig and inverse trig functions, arithmetic operations). There are closed-form formulas involving more advanced functions such as the digamma function, but those functions are essentially just defined to be equal to the sum.

EDIT: OP has no clue what to ask. If it's a product, write out a few terms, and see what cancels. Next time, be a little more careful, please.

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    Sorry, I messed up - it's a product from 2 to infinity, as in 3/4 * 8/9 * 15/16..., not a sum. Sorry about that!2012-05-29