Infinite Product Problem

Question

I need some help with this infinite product (or rather limit of partial products) which really gives me a headache. It goes like that:

$$\lim_{n \to \infty} n\prod_{m=1}^{n} \biggl(1 - \frac{1}{m} + \frac{5}{4m^2}\biggr)$$

Rewriting this gave me a term which I think is much easier to work with: $$\lim_{n \to \infty} n\prod_{m=1}^{n} \biggl(\frac{4m^2 - 4m + 5}{4m^2}\biggr)$$

Then I saw that the "5" in the numerator is what really causes a lot of problems. Because changing the "5" with a "1" gave me a fairly easy to solve product:

\begin{align} L &= \lim_{n \to \infty} n\prod_{m=1}^{n} \biggl(\frac{4m^2 - 4m + 1}{4m^2}\biggr) = \lim_{n \to \infty} n\prod_{m=1}^{n} \biggl(\frac{(2m-1)^2}{(2m)^2}\biggr) \\ &= \lim_{n \to \infty} n\prod_{m=1}^{n} \biggl(\frac{(2m-1)^2}{(2m)^2} \frac{2m+1}{2m+1}\biggr) = \lim_{n \to \infty} n\prod_{m=1}^{n} \biggl(\frac{(2m-1)(2m+1)}{(2m)^2} \frac{2m-1}{2m+1}\biggr) \\ &= \lim_{n \to \infty} n\prod_{m=1}^{n} \biggl(\frac{(2m-1)(2m+1)}{(2m)^2} \biggr) \biggl(\prod_{m=1}^{n} \frac{2m-1}{2m+1}\biggr) = \lim_{n \to \infty} n\prod_{m=1}^{n} \biggl(\frac{(2m-1)(2m+1)}{(2m)^2} \biggr)\cdot \frac{1}{2n+1} \\ &= \lim_{n \to \infty} \frac{n}{2n+1} \prod_{m=1}^{n} \biggl(\frac{(2m-1)(2m+1)}{(2m)^2} \biggr) \\ &= \frac{1}{2}\cdot\frac{2}{\pi} = \frac{1}{\pi}, \end{align} where I used the expression for the Wallis product in the last equality (and in fact the Wallis product is given as a hint).

I'd like to think that this observation has something to do with the solution. However, I couldn't come up with anthing meaningful so far. I just noticed that we can rewrite the original problem like that:

$$\lim_{n \to \infty} n\prod_{m=1}^{n} \biggl(\frac{4m^2 - 4m + 5}{4m^2}\biggr) = \lim_{n \to \infty} n\prod_{m=1}^{n} \biggl(\frac{4m^2 - 4(m-1) + 1}{4m^2}\biggr),$$ but I couldn't really make use of that. Could maybe somebody help me to continue with the Problem? Or is my approach entirely wrong and there is a much better approach for tackling this problem?

Answer 1

Consider if you were to write the infinite polynomial expansion of $\ sinx$. We know that $\ sinx$ is 0 for $\ m\pi$ and thus they all behave as roots of the polynomial , thus we can write

$$\ sinx = A\cdot x \cdot (x-\pi)\cdot (x+\pi)\cdot (x-2\pi)\cdot (x+2\pi)\cdot (x-3\pi)\cdot (x+3\pi)......$$

$$\ sinx = A \cdot x \cdot (x^2-\pi^2) \cdot (x^2-4\pi^2)......$$

To determine A, consider the limit of the infinite product of $ \frac{\ sinx}{x}$ approaches 1 as $\ x$ tends to $\ 0$.

Thus, we obtain $\ A = \prod_{m=1}^{\infty}(-\pi^2m^2)$

Substituting this $\ A$ back into our expression, we get:

$$\ sinx = x \cdot \prod_{m=1}^{\infty}(1-\frac{x^2}{m^2\pi^2})$$ This is a fun product in itself, but for now, we are more interested in the product for $\ cosx$.

Using $\ 2sinxcosx = sin2x$, we get

$$\ cos(x) = \frac{2x \cdot \prod_{m=1}^{\infty}(1-\frac{4x^2}{m^2\pi^2})}{2 \cdot x \cdot \prod_{m=1}^{\infty}(1-\frac{x^2}{m^2\pi^2})}$$

Split the even and odd products in the numerator and simplify. You will observe that it cancels out nicely and gives us the far better result:

$$\ cos(x) = \prod_{m=1}^\infty\left(1-\frac{4x^2}{\pi^2(2m-1)^2}\right)$$

Replace $\ x$ with $\ \pi x$ and we get:

$$\cos(\pi x) = \prod_{m=1}^\infty\left(1-\frac{4x^2}{(2m-1)^2}\right)$$

Consider the expression when $x=i$.

Note that $$\ cos(i\pi) = \frac{e^{i(i\pi)} + e^{-i(i\pi)}}{2} = \frac{e^{-\pi} + e^{\pi}}{2} = cosh(\pi)$$

By the very definition of the hyperbolic cosine.

The expression then evolves into

$$ \cos(i\pi) = \cosh(\pi) = \prod_{m=1}^\infty\left(1+\frac{4}{(2m-1)^2}\right)$$

As you mentioned, we also have the Wallis product to help us out:

$$\prod_{m=1}^{\infty}\frac{(2m-1)(2m+1)}{(2m)^2} = \frac{2}{\pi}$$

We're almost done. Multiplying the two infinite series, we get

\begin{align} n\prod_{m=1}^n\left(1-\frac{1}{m}+\frac{5}{4m^2}\right)&=n\prod_{m=1}^n\left(\frac{(2m-1)^2+4}{(2m)^2}\right)\\ &=n\prod_{m=1}^n\left(\frac{(2m-1)^2+4}{(2m-1)^2}\right)\cdot \prod_{m=1}^n\left(\frac{(2m-1)^2}{(2m)^2}\right)\\ &=\frac{n}{2n+1}\prod_{m=1}^n\left(1+\frac{4}{(2m-1)^2}\right)\cdot \prod_{m=1}^n\left(\frac{(2m-1)(2m+1)}{(2m)^2}\right)\\ &= \frac{1}{2}\cosh(\pi)\cdot \frac{2}{\pi} = \frac{\cosh(\pi)}{\pi}\end{align}

Answer 2

The infinite product expansion of the hyperbolic cosine is $$\cosh z = \prod_{k=1}^\infty \left(1 + \frac{z^2}{(k-\frac{1}{2})^2 \pi^2}\right), \tag{1}$$

which for $z = \pi$, becomes $$\cosh \pi = \prod_{k=1}^\infty \frac{4k^2 - 4k + 5}{(2k-1)^2}. \tag{2}$$ So the finite product part of your expression is $$S_n = \prod_{m=1}^n \frac{4m^2 - 4m + 5}{4m^2} = \prod_{m=1}^n \frac{4m^2 - 4m + 5}{(2m-1)^2} \left(\frac{2m-1}{2m}\right)^2. \tag{3}$$ And now we can use the calculation of $L$ that you performed earlier: $$\lim_{n \to \infty} n S_n = L \lim_{n \to \infty} \prod_{m=1}^n \frac{4m^2 - 4m + 5}{(2m-1)^2} = \frac{\cosh \pi}{\pi}, \tag{4}$$ where $L = \frac{1}{\pi}$. Note you have an extra factor of $2$ because $$\prod_{m=1}^n \frac{2m-1}{2m+1} = \frac{1}{2n+1}.$$