For the series $\sum_{k=1}^{\infty}a_k$, suppose that there is a number $r$ with $0\leq r<1$ and a natural number $N$ such that $$|a_k|^{1/k}<r\qquad\text{for all indices $k\geq N$}$$ Prove that $\sum_{k=1}^{\infty}a_k$ converges absolutely.
Proof:
For a given $r\in\mathbb{R}$ with $0\leq r<1$ and $N\in\mathbb{N}$ satisfy $|a_k|^{1/k}<r$ for all indices $k\geq N$, that gives $|a_k|<r^k$. Now, define $s_n=\sum_{k=1}^{n}|a_k|$ be a sequence of partial sum of $\sum_{k=1}^{\infty}|a_k|$. Since $\sum_{k=1}^{n}r^k$ converges to $(1-r^{n+1})/(1-r)$, for all $\epsilon>0$, this gives $$\left|\sum_{k=1}^{n}r^k-\frac{1-r^{n+1}}{1-r}\right|<\frac{\epsilon}{2}\qquad\text{for all $k\geq N$}$$ Then for all $j,k\geq N$, we have \begin{align*} \left|\sum_{j=1}^{n}a_j-\sum_{k=1}^{n}a_k\right|<\left|\sum_{j=1}^{n}r^j-\sum_{k=1}^{n}r^k\right|&=\left|\sum_{j=1}^{n}r^j-\frac{1-r^{n+1}}{1-r}+\frac{1-r^{n+1}}{1-r}-\sum_{k=1}^{n}r^k\right|\\ &\leq\left|\sum_{j=1}^{n}r^j-\frac{1-r^{n+1}}{1-r}\right|+\left|\frac{1-r^{n+1}}{1-r}-\sum_{k=1}^{n}r^k\right|\\ &=\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon \end{align*} Hence, $\{s_n\}$ is a Cauchy sequence which implies $\{s_n\}$ is convergent, so there exists an $M\in\mathbb{R}$ such that $\sum_{k=1}^{n}a_k\leq M$. This inequality implies $\sum_{k=1}^{\infty}|a_k|$ is convergent; therefore, $\sum_{k=1}^{\infty}a_k$ converges absolutely.
Does this solution valid? If not, can someone give me a hint or suggestion to receive the answer? Thanks.
If you know the comparison test, then you can prove the proposition easily. You know $|a_k| \le r^k$ for all $k \ge N$. Since $0\le r < 1$, the geometric series $\sum_{n=1}^{\infty} r^n$ converges. Therefore, by comparison test, $\sum_{n=1}^{\infty} |a_n|$ converges.
The proposition you have to prove is called the 'root test'.