Monster Limit 3

Suppose,

\(\displaystyle f(x) = \sum_{n=0}^{\infty} f_n x^n\), \(\displaystyle F(n) = \sum_{r=0}^{n} f_{r}\) and \(\displaystyle \lim_{n \to \infty} \dfrac{F(n)}{n^{r_f}} = \text{d}_f \)

Similarly,

\(\displaystyle g(x) = \sum_{n=0}^{\infty} g_n x^n\), \(\displaystyle G(n) = \sum_{r=0}^n g_r\) and \(\displaystyle \lim_{n \to \infty} \dfrac{G(n)}{n^{r_g}} = \text{d}_g \)

From Monster Limit 2, we have,

\[\begin{cases} \lim_{x \to 1^{-}} (1-x)^{r_f} \ f(x) = \text{d}_f \ \Gamma(1+r_f)\\ \lim_{x \to 1^{-}} (1-x)^{r_g} \ g(x) = \text{d}_g \ \Gamma(1+r_g) \end{cases}\]

Now, consider the same quantities for \(fg(x) = f(x)g(x)\), i.e,

\(\displaystyle fg(x) = \sum_{n=0}^{\infty} {fg}_n x^n\), \(\displaystyle FG(n) = \sum_{r=0}^{n} {fg}_{r}\) and \(\displaystyle \lim_{n \to \infty} \dfrac{FG(n)}{n^{r_{fg}}} = \text{d}_{fg} \)

Again,

\(\displaystyle \lim_{x \to 1^{-}} (1-x)^{r_{fg}} \ fg(x) = \text{d}_{fg} \ \Gamma(1+r_{fg})\)

Prove That
\[\displaystyle \text{d}_{fg} = \text{d}_{f} \text{d}_{g} \dfrac{\Gamma(1+r_f) \Gamma(1+r_g)}{\Gamma(1+r_f + r_g)}\]

Note by Ishan Singh
2 weeks, 5 days ago

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