# A question about convergence interval of power series

Could you give me some hint how to solve this problem:

Suppose $a_n$ is sequence defined as $a_1=\frac12,a_{n+1}=\frac12\left({a_n}^2+a_n\right)$. I managed to prove that $a_n$ is decreasing sequence, $a_n\to0$ and radius of convergence of power series $\sum_{n\ge1}a_nx^n$ is 2.

This is all quite standard staff, but how to find convergence interval ?

I could not decide about convergence of $\sum_{n\ge1}2^na_n$ because: 1)ratio test is inconclusive; 2)root test=$2\sqrt[n]{a_n}$ and I could not estimate $\sqrt[n]{a_n}$; 3)I tried comparison test,knowing that $\sum_{n\ge1}2^na_{2^n}$ but could not compute the $\lim_{n\to\infty}\frac{a_n}{a_{2^n}}$, $a_n$ is decreasing, therefore from some n$a_n\ge a_{2^n}$ but does $\frac{a_n}{a_{2^n}}$ converge to finite number ?

Thanks.

Hint: Note that

$$a_{n + 1} > \frac 1 2 a_n$$

for every $n$, so a brief argument (perhaps with induction) shows that

$$a_{n + 1} > \frac 1 {2^n} a_n$$

Hence $2^{n + 1} a_{n + 1} > 2 a_n$. Consider something similar at $-2$.

• Could you please add more how you last statement helps conclude about convergence ? – user97484 May 4 '14 at 5:15

Hint: The radius of convergence is defined as $1/\beta$, where $\beta = \lim\sup(|a_n|^{1/n})$.

• So $lim_{n\to \infty}\sqrt[n]{a_n}\le limsup{\sqrt[n]{a_n}}=\frac12$ and thus $2\lim_{n\to \infty}\sqrt[n]{a_n}\le 1$. This is still inconclusive. – user97484 May 4 '14 at 7:47