Question

For any \(s \in E^{1}\) and \(n \in \bar{N},\) define $$ \left(\begin{array}{l} s \\ n \end{array}\right)=\frac{s(s-1) \cdots(s-n+1)}{n !} \text { with }\left(\begin{array}{l} s \\ 0 \end{array}\right)=1 \text { . } $$ Then prove the following. (i) \(\lim _{n \rightarrow \infty} n\left(\begin{array}{l}s \\\ n\end{array}\right)=0\) if \(s>0\). (ii) \(\lim _{n \rightarrow \infty}\left(\begin{array}{l}s \\\ n\end{array}\right)=0\) if \(s>-1\). (iii) For any fixed \(s \in E^{1}\) and \(x \in(-1,1),\) $$ \lim _{n \rightarrow \infty}\left(\begin{array}{l} s \\ n \end{array}\right) n x^{n}=0 $$ hence $$ \lim _{n \rightarrow \infty}\left(\begin{array}{l} s \\ n \end{array}\right) x^{n}=0 $$

Short answer

The limits are zero in all cases due to the factorial growth in the denominator.

Step-by-step solution

Understand the Given Expression

We are given the expression \( \binom{s}{n} = \frac{s(s-1) \cdots (s-n+1)}{n!} \), which represents the generalized binomial coefficient. Note that \( \binom{s}{0} = 1 \) by definition.

Prove Part (i): Limit of n\binom{s}{n}

For \( s>0 \), consider the limit \( \lim_{n \rightarrow \infty} n\binom{s}{n} \). It evaluates to \( \lim_{n \rightarrow \infty} \frac{n \cdot s(s-1) \cdots (s-n+1)}{n!} \). Observe that as \( n \rightarrow \infty \), there are \( n \) factors in the numerator but \( n! = n(n-1)(n-2)\cdots(1) \) grows faster than the product in the numerator since it has \( n \) terms which grow past the fixed terms from \( s \). Hence, the numerator grows slowly compared to the denominator and the limit approaches zero.

Prove Part (ii): Limit of \binom{s}{n}

For \( s > -1 \), consider the limit \( \lim_{n \rightarrow \infty} \binom{s}{n} \). Using a similar argument as in Part (i), the terms \( (s-n+1) \) through \( 1 \) are included in the numerator, but the denominator \( n! \) still grows much more rapidly for large \( n \). Thus, \( \lim_{n \rightarrow \infty} \frac{s(s-1) \cdots (s-n+1)}{n!} = 0 \).

Prove Part (iii): Limit of \binom{s}{n} n x^n

Using Parts (i) and (ii), observe that \( \lim_{n \rightarrow \infty} \binom{s}{n} n = 0 \) leads to \( \lim_{n \rightarrow \infty} \binom{s}{n} n x^n = 0 \) for any fixed \( s \in E^1 \) and \( x \in (-1,1) \). The exponential decay of \( x^n \) as \( n \) increases outweighs any increment incurred by multiplying with \( n \binom{s}{n} \).

Conclusion for Part (iii): Limit of \binom{s}{n} x^n

It follows from the previous steps that \( \lim_{n \rightarrow \infty} \binom{s}{n} x^n = 0 \) since the product \( \binom{s}{n} n x^n \) converging to zero implies \( \binom{s}{n} x^n \) converges to zero as \( n \rightarrow \infty \).

Key concepts

Limit of Sequences

When we talk about the limit of sequences, we are exploring how a sequence behaves as it progresses towards infinity. Imagine stacking numbers one after the other, watching where they are headed as the stack grows taller. In many cases, each step of a sequence is calculated using a formula, and we become curious about what value it will approach as the number of steps increases without bound.

The problem provided explores limits in three different scenarios. For instance, in part (i), we examine the behavior of the expression \( n \binom{s}{n} \) as \( n \) increases, where \( s > 0 \). Intuitively, despite the presence of \( s \) terms in the numerator, the denominator grows larger much faster, like a race where one runner eventually always falls behind.

When we mathematically declare \( \lim_{n \to \infty} f(n) = L \), it means as \( n \) gets almost unimaginably large, the elements \( f(n) \) of our sequence get arbitrarily close to \( L \). Limits increase our understanding of infinite processes, helping us predict and make conclusions about sequences beyond our scope of direct computation.

Factorials

Factorials, denoted by an exclamation mark, are a fascinating concept offering a blend of simplicity and power. Written as \( n! \), it represents the product of all positive integers from 1 to \( n \). So, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

In our problem, factorials form the denominator of the generalized binomial coefficient \( \binom{s}{n} = \frac{s(s-1) \dots (s-n+1)}{n!} \). This structure is fundamental because factorials grow exceptionally fast, faster than most polynomial expressions like the one in the numerator. Consider how each successive product in \( n! \) is larger than the last: 2 is larger than 1, 3 is larger than 2, and so on. This compounding factor results in exponential growth behavior.

Factorials are pivotal in understanding why, with increasing \( n \), the expression \( \binom{s}{n} \) tends to diminish towards zero in parts (ii) and (iii) of our problem. The rapid growth of the factorial dwarfs the changes in the polynomial, which is why terms like \( n \cdot x^n \) ultimately don't upset this delicate balance.

Convergence of Series

Convergence is a concept that tells us how a series behaves when all its terms are summed up over an infinite horizon. We say a series is convergent if it approaches a specific value, or "converges", as more of its terms are considered. For instance, if the terms of a series \( a_n \) become smaller and cause the series \( \sum a_n \) to settle at a point, that series converges.

In part (iii) of the given problem, we investigate a series formed not just by the terms \( \binom{s}{n} x^n \) but also influenced by the multiplier \( n \). Here, \( x^n \) becomes critical as \( x^n \) \text{diminishes to zero\ for any \(-1 < x < 1\). Imagine water droplets evaporating faster than they can accumulate; tiny droplets (elements) keep vanishing, leading the sum to zero.

The convergence observed in our exercise underscores how individual elements—while growing larger initially or experiencing fluctuations over time—cannot subvert this comprehensive push towards zero. Hence, understanding convergence helps us not only solve but also predict outcomes of infinite sums, preparing us for more complex mathematical challenges ahead.