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**HTale** Hi guys,

I'm having real difficulty with this

Show that

$\displaystyle \lim_{k\to\infty} \prod^k_{r=1}\left(1 - \frac{\sin^2\left(\frac{\pi x}{n}\right)}{\sin^2\left(\frac{\pi r}{n}\right)}\right) = \lim_{k\to\infty}\prod^{k}_{r=1} \left( 1 - \frac{x^2}{r^2} \right)$, where $\displaystyle n = 2k+1$.

Now, I know by L'Hopital's rule, that

$\displaystyle \lim_{n\to\infty}\left(1 - \frac{\sin^2\left(\frac{\pi x}{n}\right)}{\sin^2\left(\frac{\pi r}{n}\right)}\right) = 1 - \frac{x^2}{r^2} $

But the relation of n with k has really thrown me off.

Any help would be much appreciated.

Thanks in advance,

HTale.

Your problem amounts to interchanging a limit and an infinite product (or a series), can you see why?

You need to show $\displaystyle \lim_k \prod_{r=1}^k \left(1 - \frac{\sin^2\left(\frac{\pi x}{2k+1}\right)}{\sin^2\left(\frac{\pi r}{2k+1}\right)}\right) = \prod_{r=1}^\infty \left(1 - \frac{x^2}{r^2}\right) $. Due to your remark, this can be rewritten as

$\displaystyle \lim_k \prod_{r=1}^\infty a(r,k) = \prod_{r=1}^\infty \lim_k a(r,k)$

where $\displaystyle a(r,k)=1 - \frac{\sin^2\left(\frac{\pi x}{2k+1}\right)}{\sin^2\left(\frac{\pi r}{2k+1}\right)}$ if $\displaystyle r=1,\ldots,k$ and $\displaystyle a(r,k)=1$ if $\displaystyle r>k$.

You need a theorem to justify that. A good one would be the "dominated convergence theorem" (you may know another name):**Dominated convergence theorem (for series): **

If $\displaystyle (u(r,k))_{r,k\geq 0}$ is such that

- for every $\displaystyle r$, $\displaystyle u(r,k)\to_{k\to\infty} u(r)$;
- there is $\displaystyle (v(r))_{r\geq 0}$ such that $\displaystyle |u(r,k)|\leq v(r)$ for every $\displaystyle r,k$, and $\displaystyle \sum_{r=0}^\infty v(r)<\infty$,

Then, for every $\displaystyle k$, the series $\displaystyle \sum_r u(r,k)$ and $\displaystyle \sum_r u(r)$ converge, and we have $\displaystyle \lim_{k\to\infty} \sum_{r=0}^\infty u(r,k)=\sum_{r=0}^\infty u(r)$.

I wrote the theorem for series, so you must let $\displaystyle u(r,k)=\log a(r,k)$ in order to have $\displaystyle \sum_r u(r,k)=\log\prod_r a(r,k)$ and apply the theorem.

Thus you must find $\displaystyle v(r)\geq 0$ such that $\displaystyle \sum_r v(r)$ converges and, for every $\displaystyle 1\leq r\leq n$,

$\displaystyle \left|\log\left(1 - \frac{\sin^2\left(\frac{\pi x}{2k+1}\right)}{\sin^2\left(\frac{\pi r}{2k+1}\right)}\right)\right|\leq v(r)$.

(And the inequality $\displaystyle |u(r,k)|\leq v(r)$ is obvious if $\displaystyle r>n$ since we would have $\displaystyle u(r,k)=0$)

To this aim, you can use the following bounds: if $\displaystyle 0\leq u\leq \frac{\pi}{2}$, then $\displaystyle \frac{2}{\pi}u\leq \sin u\leq u$. I let you carry on from there.

Note: I assumed $\displaystyle 0<x<1$ so that the factors of the product are positive and the logarithm makes sense.