Suppose $\sum_{n=1}^\infty \epsilon_n$ converges.
I know the result that $\sum_{n=1}^\infty \epsilon_n^p$ converge for $p \ge 1,$ and I try to use the root test to prove it.
Proof : $\sum_{n=1}^\infty \epsilon_n \lt \infty \Longleftrightarrow \lim_{n\to \infty} \sqrt[n]{\epsilon_n} \lt 1 \Longleftrightarrow \lim_{n\to \infty} \sqrt[n]{\epsilon_n^p} \lt 1 \text{ (Since $x^p$ is continuous function)} \Longleftrightarrow \sum_{n=1}^\infty \epsilon_n^p \lt \infty.$
End of Proof.
But I don't see why the proof using the root test fails in the case $0\lt p \lt 1$.
The series $\sum_{n=1}^\infty \epsilon_n^p$ diverge when $0\lt p \lt 1,$ right?
I am so confused and sincerely want some help. Thanks in advance.
This is wrong. It is not true that$$\sum_{n=1}^\infty \varepsilon_n < \infty \Longleftrightarrow \lim_{n\to \infty} \sqrt[n]{\varepsilon_n} < 1.$$For instance, take $\varepsilon_n=\frac1{n^2}$. In this case, you have$$\sum_{n=1}^\infty \varepsilon_n < \infty \text{ and } \lim_{n\to \infty} \sqrt[n]{\varepsilon_n} = 1.$$The root test only asserts that$$\sum_{n=1}^\infty \varepsilon_n < \infty \Longleftarrow \lim_{n\to \infty} \sqrt[n]{\varepsilon_n} < 1.$$
Assuming that $(\forall n\in\mathbb{N}):\varepsilon_n\geqslant0$, you can solve your problem as follows: since $\sum_{n=1}^\infty\varepsilon_n<+\infty$, $\varepsilon_n<1$ if $n$ is large enough. But ${\varepsilon_n}^p\leqslant\varepsilon_n$. Therefore, you can apply the comparison test.