Consider the power series
$$\sum_{n=1}^\infty \frac{(-1)^n}{n} z^{n(n+1)}$$
and determine its convergence radius $R$.
Attempt:
Write the coefficient sequence of the power series as $(a_n)_{n=1}^\infty$, we are interested in determining $$R = 1/\limsup_{n \to \infty} |a_n|^{1/n}$$
We have
$$(a_n)_n = \left(0,0,-1,0,0,0,1/2,0,0,0,0,0,\frac{-1}{3}, \dots\right)$$
and thus $(|a_n|^{1/n}) = \left(0,0,1^{1/2},0,0,0,(1/2)^{1/6}, …\right)$
and we see that $|a_n|^{1/n}$ contains the subsequence
$$\frac{1}{n^{1/[n(n+1)]}}$$
This subsequence converges to $1$. This can be seen by taking logarithms and the well known limit $n^{1/n} \to 1$. Hence, it follows that $\limsup_n |a_n|^{1/n} \geq 1$. Since all terms in the subsequence are smaller than $1$, the limsup of the sequence is smaller than $1$. Hence, the limsup is precisely $1$ and $R= 1/1 = 1.$
Is this correct?
Best Answer
Yes this is correct.
Another solution could have been: The series converges for $z=1$ (decreasing alternating series). This implies $R\geq1$.
And it diverges for $|z|>1$ since then $\frac{( -1)^nz^{n(n+1)}}{n}\not\to0.$ This implies $R\leq1$. Therefore $R=1.$