1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $-2$.

$\endgroup$
  • $\begingroup$ Could you please add more how you last statement helps conclude about convergence ? $\endgroup$ – user97484 May 4 '14 at 5:15
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user97484 May 4 '14 at 7:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.