Radius of convergence of $\sum(n-1/n)^{n^2}z^n$

Question

$$\sum_{n=0}^\infty {(n-1/n)^{n^2} z^n}$$ Can someone help me with this exercise? I've applied ratio test for convergence, but I haven't been able to menage with the power of power.

Answer 1

Let $a_n:=(n-1/n)^{n^2}$ and show that $a_n^{1/n} \ge \frac{n^n}{2^n}$.

Conclusion ?

Answer 2

$a_n=n^{n^2}(1-\frac {1}{n^2})^{n^2}$. You can consider the nth root of this and apply the result on radius of convergence for power series.

Answer 3

Let $$a_n=\left(n-\frac{1}{n}\right)^{n^2}\implies \log(a_n)=n^2\log\left(n-\frac{1}{n}\right)$$ Now $$\log(a_{n+1})-\log(a_n)=(n+1)^2 \log\left(n+1-\frac{1}{n+1}\right)-n^2\log\left(n-\frac{1}{n}\right)$$ Now, rewriting $x-\frac 1x=x\left(1-\frac 1{x^2}\right)$, use Taylor expansion to get $$\log(a_{n+1})-\log(a_n)=n \left(1+2 \log \left({n}\right)\right)+\left(\frac{3}{2}+\log \left({n}\right)\right)+\frac{1}{3 n}+O\left(\frac{1}{n^2}\right))$$ and then $$\frac{a_{n+1} }{a_n}=e^{ \log(a_{n+1})-\log(a_n)}$$