Question

(a) Show that $$ \lim _{n \rightarrow \infty}\left(1-\frac{\alpha}{1}\right)\left(1-\frac{\alpha}{2}\right) \cdots\left(1-\frac{\alpha}{n}\right)=0, \quad \text { if } \quad \alpha>0 $$ (b) Conclude from (a) that $$ \lim _{n \rightarrow \infty}\left(\begin{array}{l} q \\ n \end{array}\right)=0 \quad \text { if } \quad q>-1 $$ where \(\left(\begin{array}{l}q \\ n\end{array}\right)\) is the generalized binomial coefficient of Example 2.5.3.

Short answer

Solution: The limit of the given expression as \(n\) approaches infinity for \(\alpha > 0\) is 0. Explanation: By simplifying the expression, taking the natural log of the expression, and using various approximations (log approximation, harmonic series), we established that the limit of the product expression as \(n\) approaches infinity is 0 for \(\alpha > 0\).

Step-by-step solution

Simplify the given expression for the limit

The given expression is: $$ \lim _{n \rightarrow \infty}\left(1-\frac{\alpha}{1}\right)\left(1-\frac{\alpha}{2}\right) \cdots\left(1-\frac{\alpha}{n}\right) $$ Let's see if we can simplify this expression a bit by noticing that it is equivalent to: $$ \lim _{n \rightarrow \infty}\prod_{k=1}^{n} \left(1-\frac{\alpha}{k}\right) $$ which is the limit of a product as \(n\) approaches infinity.

Taking log of the expression

Let's take the natural log of the expression inside the limit: $$ \ln\left(\prod_{k=1}^{n} \left(1-\frac{\alpha}{k}\right)\right) = \sum_{k=1}^{n}\ln\left(1-\frac{\alpha}{k}\right) $$ By taking the natural log, we have converted the product into a sum.

Use log approximation

We will now use the fact that \(\ln(1+x) \approx x\) for small values of \(x\), which is true because of the first-order approximation of the Taylor series for the natural log function. Applying this approximation, we get: $$ \sum_{k=1}^{n}\ln\left(1-\frac{\alpha}{k}\right) \approx \sum_{k=1}^{n} -\frac{\alpha}{k} = -\alpha\sum_{k=1}^{n} \frac{1}{k} $$

Use the harmonic series approximation

The series \(\sum_{k=1}^{n} \frac{1}{k}\) is known as the harmonic series. This series diverges as \(n\) approaches infinity, and its behavior is approximately \(\ln(n)\). Thus, we can write: $$ -\alpha\sum_{k=1}^{n} \frac{1}{k} \approx -\alpha\ln n $$

Find the limit

Now we can find the limit of the original expression: $$ \lim _{n \rightarrow \infty}\prod_{k=1}^{n} \left(1-\frac{\alpha}{k}\right) = \lim_{n \rightarrow \infty} e^{-\alpha\ln n} = \lim_{n \rightarrow \infty} n^{-\alpha} $$ For \(\alpha>0\), we have that the limit is 0: $$ \lim_{n \rightarrow \infty} n^{-\alpha} = 0 $$

Connect to the generalized binomial coefficient

Now we need to use the result from part (a) to derive the limit for the generalized binomial coefficient: $$ \lim _{n \rightarrow \infty}\left(\begin{array}{l} q \\\ n \end{array}\right)=0 \quad \text { if } \quad q>-1 $$ Recall from Example 2.5.3 that the generalized binomial coefficient is given by: $$ \left(\begin{array}{l} q \\\ n \end{array}\right) = \frac{\Gamma(q+1)}{\Gamma(n+1)\Gamma(q-n+1)} $$ where \(\Gamma(x)\) is the Gamma function. For large values of \(n\), we can use the Stirling approximation for the Gamma function, which is: $$ \Gamma(x+1) \approx x^x e^{-x}\sqrt{2\pi x} $$ Using this approximation, we have that: $$ \left(\begin{array}{l} q \\\ n \end{array}\right) \approx \frac{q^q e^{-q}\sqrt{2\pi q}}{n^n e^{-n}\sqrt{2\pi n}(q-n)^{(q-n)} e^{-(q-n)}\sqrt{2\pi (q-n)}}\cdot\frac{1}{\sqrt{2\pi}} $$ Simplifying the expression, we get: $$ \left(\begin{array}{l} q \\\ n \end{array}\right) \approx \frac{q^q}{n^n (q-n)^{(q-n)}}e^{n} $$ As \(n\) goes to infinity, we observe that the second term \(n^n\) in the denominator will dominate the others, and thus the limit will tend to 0, which matches the result in part (a). Therefore, we have shown that: $$ \lim _{n \rightarrow \infty}\left(\begin{array}{l} q \\\ n \end{array}\right)=0 \quad \text { if } \quad q>-1 $$

Key concepts

Convergence of Infinite Products

Exploring the convergence of infinite products is vital in real analysis, particularly when understanding complex functions and series. The product
\[\begin{equation}\textstyle\biggl(\textstyle\biggl(1-\frac{\alpha}{1}\textstyle\biggr)\textstyle\biggl(1-\frac{\alpha}{2}\textstyle\biggr)\cdots\textstyle\biggl(1-\frac{\alpha}{n}\textstyle\biggr)\biggr)\forall \alpha > 0\end{equation}\]can be seen as an infinite product as n approaches infinity. To assess its convergence, one can utilize the properties of logarithms, transforming the product into a sum through the natural logarithm function.
This process showcases a sequence of partial products and their behavior as n grows without bounds. If, as in our exercise, the limit of the sequence of partial products approaches zero, we say that the infinite product converges to zero. This notion is fundamental when dealing with complex functions wherein each factor of the product can affect the overall behavior of the series or function.

Generalized Binomial Coefficient

The generalized binomial coefficient, denoted as
\(\left(\begin{array}{l}q \n\end{array}\right)\),extends the concept of binomial coefficients to real and complex indices. In contrast to the classic binomial coefficients, which are defined only for non-negative integer values, the generalized binomial coefficient includes cases where q can be any real or complex number. This is particularly useful for power series expansions and probability theory.
Using the Gamma function, we can express the generalized binomial coefficient for any real number q, which allows us to explore a broader range of problems. As the exercise indicates, the limit of the generalized binomial coefficient as n approaches infinity is zero for any q greater than -1, emphasizing the coefficient's behavior at the boundaries of its definition.

Harmonic Series Divergence

The harmonic series is the sum of reciprocals of the natural numbers, expressed as
\(\sum_{k=1}^{\infty} \frac{1}{k}\). Despite its simplicity, it exhibits a curious property: it diverges, meaning it grows without bound as more terms are added.
This divergence is illustrated through the natural logarithmic approximation, where the partial sums of the harmonic series are asymptotically equivalent to the natural logarithm of n, or \(\ln(n)\), as n grows large. Understanding the divergence of the harmonic series is imperative in real analysis, since it helps in the assessment of series convergence and aids in various approximation techniques. It also helps explain why our initial infinite product tends to zero as we involve the terms of the harmonic series in our calculation.

Stirling's Approximation

Stirling's approximation is a powerful tool in mathematics, particularly in the realms of probability, statistics, and real analysis. It serves as an approximate formula for the factorial function, which is essential in computing the probabilities of complex events as well as in the analysis of large-scale behaviors. Stirling's approximation is given by
\(\Gamma(x+1) \approx x^x e^{-x}\sqrt{2\pi x}\),
where \(\Gamma(x)\) is the Gamma function — an extension of the factorial function to real and complex numbers. In contexts where n becomes very large, such as with the generalized binomial coefficient in our exercise, Stirling's approximation provides a way to simplify the factorial term. It's an elegant asymptotic expression that, while simplifying calculations, retains significant accuracy for large n, which establishes its widespread use in applied and theoretical work alike.