Limits: Does (1 - 1/4)(1 - 1/9) ... (1 - 1/n^2) converge as n approaches zero?

[kero9kero]

I got this problem in my homework and I know it converges as n approaches infinity, but I'm stuck as it approaches zero. These are the steps I've done so far:

Statement: Find the limit.



First I factorized the equation:





And cancelling terms, I got this:


So, is there something I did wrong?

 

[MarkFL]

I would say that the expression for which we are trying to find the limit may be stated:

\(\displaystyle \displaystyle f(n)=\prod_{k=2}^n\left(1-\frac{1}{k^2}\right)\)

And so \(\displaystyle f\) is only defined for \(\displaystyle 2\le n\).

 

[JeffM]

What was the exact wording of the question? What you have stated makes no sense.

\(\displaystyle \displaystyle \left ( \prod_{j=2}^n 1 - \dfrac{1}{j^2} \right ) \implies\)

\(\displaystyle n \text { is an integer } > 1.\)

It is, however, true that

\(\displaystyle n \ge 2 \implies \displaystyle \left ( \prod_{j=2}^n 1 - \dfrac{1}{j^2} \right ) = \dfrac{n + 1}{2n}.\)

 

[Dr.Peterson]

I got this problem in my homework and I know it converges as n approaches infinity, but I'm stuck as it approaches zero. These are the steps I've done so far:

Statement: Find the limit.

View attachment 10428

First I factorized the equation:

View attachment 10429

View attachment 10430

And cancelling terms, I got this:
View attachment 10431

So, is there something I did wrong?

The expression is only defined for integer values of n (greater than 1). It makes no sense to ask for a limit as n approaches 0. Surely that was meant to be infinity.

But your work is good.

 

[JeffM]

\(\displaystyle n \ge 2 \implies \displaystyle \left ( \prod_{j=2}^n 1 - \dfrac{1}{j^2} \right ) = \dfrac{n + 1}{2n}.\)

For anyone wondering how that statement is determined, see below. (Note the OP did not say exactly how it was obtained.) Moreover, it is not the most useful form for thinking about limits.

\(\displaystyle \displaystyle \left ( \prod_{j=2}^n 1 - \dfrac{1}{j^2} \right ) = \left \{ \prod_{j=2}^n 1 - \dfrac{1}{j} \right ) *\left ( \prod_{j=2}^n 1 + \dfrac{1}{j} \right ) = \\

\displaystyle \left \{ \prod_{j=2}^n \dfrac{j - 1}{j} * \prod_{j=2}^n \dfrac{j + 1}{j} \right \}= \left \{ \left ( \prod_{j=2}^n j - 1 \right ) \div \left ( \prod_{j=2}^n j \right)^2 \right \} * \left (* \prod_{j=2}^n j + 1 \right ) = \\]

\dfrac{(n - 1)!}{(n!)^2} * \dfrac{(n + 1)!}{2} = \dfrac{1}{2} * \dfrac{(n - 1)!}{n!} * \dfrac{(n + 1)!}{n!} = \dfrac{n + 1}{2n} =\\

\dfrac{n}{2n} + \dfrac{1}{2n}= \dfrac{1}{2} + \dfrac{1}{2n}.\)

 

[pka]

I got this problem in my homework and I know it converges as n approaches infinity, but I'm stuck as it approaches zero. These are the steps I've done so far:
Statement: Find the limit.
View attachment 10428
View attachment 10429 And cancelling terms, I got this: View attachment 10431

What was the exact wording of the question? What you have stated makes no sense.
\(\displaystyle \displaystyle \left ( \prod_{j=2}^n 1 - \dfrac{1}{j^2} \right ) \implies\) \(\displaystyle n \text { is an integer } > 1.\) It is, however, true that \(\displaystyle n \ge 2 \implies \displaystyle \left ( \prod_{j=2}^n 1 - \dfrac{1}{j^2} \right ) = \dfrac{n + 1}{2n}.\)

The expression is only defined for integer values of n (greater than 1). It makes no sense to ask for a limit as n approaches 0. Surely that was meant to be infinity. But your work is good.

I had resisted posting on this thread hoping that kero9kero might realize what DA.Peterson has.
The expression approaches one half not zero.

To kero9kero: is that a correct reading of your original post?

 

[lookagain]

> > > The expression approaches one half not zero. < < <

This is consistent if the limit is supposed to be as n approaches infinity, instead of n approaching 0, which is what the OP has.