Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to solve an exercise, but i am not sure that the result i get at the end is correct...May i kindly ask you for a little help or a remark?

Find the radius of convergence of the following power series: $\sum_{n=0}^{\infty }(n^{2}+a^{n})z^{n}$ for any $z,a\in \mathbb{C}$

I use the quotient ratio: $\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty } \left | \frac{n^{2}+2n+1+aa^{n}}{n^{2}+a^{n}} \right |=\lim_{n\rightarrow \infty }\left | \frac{1+\frac{2}{n}+\frac{1}{n^{2}}+ \frac{aa^{n}}{n^{2}}}{1+\frac{a^{n}}{n^{2}}} \right | = 1 $ and then i get radius of convergence $1$.

With Cauchy-Hadamard i get $\lim sup_{n\rightarrow \infty }\sqrt{n^{2}+a^{n}}=\lim sup_{n\rightarrow \infty }\sqrt{n^{2}(1+\frac{a^{n}}{n^{2}})}=\lim sup_{n\rightarrow \infty }n\sqrt{1+\frac{a^{n}}{n^{2}}}=\infty $ and a radius of convergence $0$.

I am not sure which result is correct if any of them is correct...

Thank you in advance!

share|improve this question
1  
If $|a| > 1$, then your quotient ratio calculation is wrong. Also, in Cauchy-Hadamard you need to take the $n$'th root, not the square root. –  Yoni Rozenshein Mar 15 '13 at 15:18
1  
It is lim sup of an $n$=th root, not a square root. –  André Nicolas Mar 15 '13 at 15:20
    
@Yoni, you're right, Cauchy-Hadamard i did wrong. I am sorry. Thank you for the correction. –  Lullaby Mar 15 '13 at 15:21

1 Answer 1

up vote 1 down vote accepted

Using the ratio test seems easier than Cauchy-Hadamard in this case:

First let us assume that $|a| \le 1$. Then the major term of $n^2 + a^n$ is $n^2$, since $a^n$ is bounded by a constant ($1$). By a calculation similar to what you did, the radius of convergence is $1$.

Second, let us assume that $|a| > 1$. Then the major term of $n^2 + a^n$ is $a^n$, which grows exponentially fast. The ratio test gives us (using the fact that exponential growth is much stronger than polynomial growth)

$$\lim_{n\to\infty} \left| \frac {a_{n+1}} {a_n} \right| = \lim_{n\to\infty} \left| \frac {a^{n+1} + (n+1)^2} {a^n + n^2} \right| = \lim_{n\to\infty} \left| \frac {a + \frac {(n+1)^2} {a^n}} {1 + \frac {n^2} {a^n}} \right| = |a|$$

which means the radius of convergence is $\frac 1 {|a|}$.

share|improve this answer
    
Thank you very much! –  Lullaby Mar 16 '13 at 19:25

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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