Let $a_n$ be a power series such that:

$$a_1=\frac{1}{2}, a_{n+1}=\frac{1}{2} \cdot \left( a_n^2+a_n \right)$$

I have to find the Radius of Convergence.

This series is monotonic decreasing (proving by induction), and bounded (by $\frac{1}{2}$ and $0$).

My quesion is if I can conclude that the Radius of Convergence is $1$, since $$ \left|\lim\limits_{n\to\infty} \frac{a_n}{a_{n+1}} \right| =\left|\frac{\lim\limits_{n\to\infty} a_n}{\lim\limits_{n \to \infty} a_{n+1}}\right| =1$$

That $R=1$.

Is it correct? If not, why?

  • $\begingroup$ $a_n$ is a sequence, so how can it be a power series? $\endgroup$ – zhw. Aug 15 '17 at 22:00

You proved the sequence is monotonically decreasing and bounded below by $0$ hence converges to $L$. Thus $L = \dfrac{L^2+L}{2}\implies L^2 = L \implies L(L-1) = 0 \implies L = 0, 1 \implies L = 0$ since $L \le 1/2 < 1$. Thus $R = \left|\displaystyle \lim_{n \to \infty} \dfrac{1}{\frac{a_{n+1}}{a_n}}\right|= \dfrac{1}{\left|\displaystyle \lim_{n \to \infty} \dfrac{a_{n+1}}{a_n}\right|}= \dfrac{1}{\dfrac{1}{2}\left(\displaystyle \lim_{n \to \infty} a_n + 1\right)}= \dfrac{1}{\frac{1}{2}} = 2$

  • $\begingroup$ Great Help, thanks $\endgroup$ – Alan Aug 14 '17 at 19:43

This is not correct, because it assumes that $\lim_{n\in\mathbb N}a_n>0$. Please note that you can can deduce nothing from your approach if $\lim_{n\in\mathbb N}a_n=0$.

  • $\begingroup$ Can't I prove by induction that $a_n >0$? $\endgroup$ – Alan Aug 14 '17 at 19:37
  • $\begingroup$ @Alan Of course they're all greater than $0$! But is the limit greater than $0$? $\endgroup$ – José Carlos Santos Aug 14 '17 at 19:41

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