Monster Limit 2

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

Give that limnA(n)nr=α\displaystyle \lim_{n \to \infty} \dfrac{A(n)}{n^r} = \alpha

Prove That

limx1(1x)rf(x)=α Γ(1+r) \lim_{x \to 1^{-}} (1-x)^r f(x) = \alpha \ \Gamma(1+r)

Note by Ishan Singh
10 months ago

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