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How do I calculate for the following power serie $\sum_{k=2}^{\infty}\begin{pmatrix} k\\ 2\end{pmatrix}(z-3i)^k$ the development point $z_0\in \mathbb{C}$ and convergence radius $R\in \left [ 0,\infty \right ]$?

Maica
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2 Answers2

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Hint. One may recall the standard geometric series evaluation $$ \sum_{k=0}^{\infty } (z-a)^k=\frac1{1-(z-a)},\qquad |z-a|<1, $$ then one is allowed to differentiate twice termwise, obtaining $$ \sum_{k=0}^{\infty } k(k-1)(z-a)^{k-2}=\frac2{\left(1-(z-a)\right)^3} $$ that is

$$ \sum_{k=0}^{\infty } k(k-1)(z-a)^{k}=\frac{2(z-a)^2}{\left(1-(z-a)\right)^3},\qquad |z-a|<1. $$

Observe that $$ k(k-1)=2 \cdot \binom k2. $$

Olivier Oloa
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  • Can you explain me why is $k(k-1)=2 \cdot \binom k2$? – Maica Apr 09 '17 at 13:06
  • @Majica Sure. You have $$\binom k2=\frac{k!}{2!(k-2)!}=\frac{k\times (k-1) \times \color{red}{(k-2)!}}{2\times\color{red}{(k-2)!}}= \frac{k\times (k-1) }{2}.$$ – Olivier Oloa Apr 09 '17 at 14:30
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Directly, by Cauchy-Hadamard formula:

$$\sqrt[k]{\binom k2}=\sqrt[k]{\frac{k(k-1)}2}\xrightarrow[k\to\infty]{}1\implies$$

the convergence radius is $\;R=1\;$ , and thus it must be that $\;|z-3i|<1\;$

DonAntonio
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