Find the radius of convergence of $\sum_{n=1}^\infty (\frac{5n^3+2n^2}{3n^3+2n})z^n$

Question

Radius of convergence of $\sum_{n=1}^\infty (\frac{5n^3+2n^2}{3n^3+2n})z^n$

I know Hadamards lemma says that $\frac{1}{R}=\lim\sup \vert a_n\vert^{1/n}$

But I'm not sure how to compute $\lim\sup \vert a_n\vert^{1/n}$

I can see the $(a_n)\to 5/3$, so I believe $\lim\sup \vert a_n\vert$ should just be the limit of the sequence? But I'm not sure how to actually use the lemma to find $R$.

Answer 1

Cauchy-Hadamard theorem is often a power hammer to kill a fly.

You can just notice that the series diverges for $z=1$ as its term doesn’t converge to zero in that case. And that the modulus of its term is less than $\frac{5}{2}\vert z\vert^k$. As the series $\sum z^k$ converges for $\vert z \vert \le 1$, the radius is equal to $1$.

Answer 2

Rational functions don't matter when finding the radius of convergence since $\lim_{n \to \infty} (n^k)^{1/n} =1 $ for any real $k$.

(This follows from $\lim_{n \to \infty} n^{1/n} =1 $.)

Also note that $a^{1/n} \to 1$ for any $a > 0$.

So, in this case, the radius is 1 - this converges for $|z| < 1$.

Also, since $a_n \to 5/3, a_n^{1/n} \to 1 $.

Answer 3

At first glance I could tell that converges when $|z|<1$ ,maybe it will help you

Answer 4

Since $a_n \to 5/3$, there is $N \in \mathbb N$ such that

$$1 \le a_n \le 2$$

for $n>N.$ Hence

$$1 \le a_n^{1/n} \le 2^{1/n}$$

for $n>N.$ This gives

$$ a_n^{1/n} \to 1.$$