# Complex variable: studying convergence of series in terms of radius of a different series

Trying to solve this problem:

If the radius of convergence of the power series $$\sum_{n=0}^\infty a_n z^n$$ is R, with $0 < R < \infty$, then the radius of convergence $R_1$ of the power series $$\sum_{n=1}^\infty n^{-n} a_n z^n$$ is

A) $R_1 = 0$

B) $R_1 = e^R$

C) $R_1 = \infty$

it is from a test we did and our professor just told us the correct answer was C. Not how to solve it.(He never explains anything).

So far I know the first series converge if $\lvert z \rvert < R$ and since $$n^{-n} \leq 1 \implies \lvert a_n z^n \rvert n^{-n} \leq \lvert a_n z^n \rvert \implies \lim \lvert a_n z^n \rvert n^{-n} \leq \lim \lvert a_n z^n \rvert \implies \frac{1}{R_1} \leq \frac{1}{R}$$ and then $R \leq R_1$. But I don't know how to get $R_1 = \infty$ with only this info.

Thanks in advance.

## 1 Answer

Use the Cauchy-Hadamard Theorem (twice) on the new series.

Suppose $R$ is the radius of the old series and $R_1$ is the radius of the new series. Then, because the limit of the numerator below, exists and is finite ($1/R$), you can write:

$$\frac{1}{R_1}=\lim_{n\to\infty}\sup\sqrt[n]{\left|\frac{a_n}{n^n}\right|}=\lim_{n\to\infty}\frac{\lim\limits_{n\to \infty}\sup\sqrt[n]{|a_n|}}{n}=\lim_{n\to\infty}\frac{\frac{1}{R}}{n}=0$$

therefore $R_1=\infty$.

Just in case your prof. hasn't given you the Cauchy-Hadamard test yet, you can also use the Ratio Test for the new coefficients $b_n=n^{-n}\cdot a_n=\frac{a_n}{n^n}$. Then,

\begin{align} \frac{b_{n+1}}{b_n}&=\frac{\frac{a_{n+1}}{(n+1)^{n+1}}}{\frac{a_n}{n^n}}\\ &=\frac{n^n}{(n+1)^{n+1}}\cdot\frac{a_{n+1}}{a_n}\\ &=\frac{1}{n+1}\cdot\frac{1}{(1+\frac{1}{n})^n}\cdot\frac{a_{n+1}}{a_n}\\ &\to 0\cdot e^{-1}\cdot 1=0 \end{align}

hence $R_1=\infty$, as above.