Decide for which x the series $\sum_{k=1}^{\infty}( \sqrt[k]{e}-1) {x^k} $ converges

Question

I want to find radius of convergence for the power series $\sum_{k=1}^{\infty}( \sqrt[k]{e}-1) {x^k}$. I try with the ratio test and I get

$\frac{|(e^{1/(k+1)}-1)x^{k+1}|}{|(e^{1/k}-1)x^{k}|} = |x| \frac{e^{1/(k+1)}}{e^{1/k}} \frac{1-1/e^{1/(k+1)}}{1-1/e^{1/k}} \longrightarrow |x| \frac{e^{0}}{e^{0}} \frac{1-1/e^{0}}{1-1/e^{0}} $ when $ k \longrightarrow \infty $

That is, I end up with a $\frac{0}{0}$ expression. I can "see" on the final expression that it approaches $|x|\times \frac{1}{1} \times \frac{1}{1}$. Is there any way to achieve this without ending up with the zero division?

Appreciates all help I can get.

Answer 1

For any $a > 1$, consider $f(a) =\sum_{k=1}^{\infty}( \sqrt[k]{a}-1) {x^k} $.

Since $\sqrt[k]{a} =e^{\ln(a)/k} \approx 1+ \ln(a)/k+O((1/k)^2) $, we have $\sqrt[k]{a}-1 \approx \ln(a)/k+O((1/k)^2) $ so the sum behaves like $\sum_{k=1}^{\infty}\dfrac{x^k\ln(a)}{k} $ and this converges for $|x| < 1$ and diverges for $|x| > 1$.

Answer 2

Let $ R_{a} $ be the radius of convergence of $ \sum{\left(\sqrt[n]{\mathrm{e}}-1\right)x^{n}} $, and $ R_{b} $ the radius of convergence of $ \sum{\frac{1}{n}x^{n}} \cdot $

Since $ \mathrm{e}^{\frac{1}{n}}-1\underset{n\to +\infty}{\sim}\frac{1}{n} $, we have $ R_{a}=R_{b} \cdot $

Observe that at the point $ x=1 $, the series diverges using the privious comparaison, and that at point $ x=-1 $ the series converges by the alternating series test.

Thus, the series converges if and only if $ x\in\left[-1,1\right[ \cdot $