Monster Limit 2

Define $\displaystyle f(x) = \sum_{n=0}^{\infty} a_{n} x^n$ and $\displaystyle A(n) = \sum_{r=0}^{n} a_{r}$

Given that $\displaystyle \lim_{n \to \infty} \dfrac{A(n)}{n^r} = \alpha$

Prove That

$\lim_{x \to 1^{-}} (1-x)^r f(x) = \alpha \ \Gamma(1+r)$

#Calculus

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`2 \times 3`	$2 \times 3$
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Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$