# Prove that $\sum_{k=1}^{\infty}a_k$ converges absolutely.

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.

• I get confused because you wrote $\sum_{j=1}^n a_j - \sum_{k=1}^n a_k$. It is $0$. Shouldn't it be corrected as $\sum_{i=N}^j a_i - \sum_{i=N}^k a_i$? – choco_addicted Feb 22 '16 at 5:06
• @choco_addicted that should be $\sum_{j=N}^{n}a_j-\sum_{k=N}^{n}a_k$ ? – Simple Feb 22 '16 at 5:17
• You have the partial sum $$s_n = \sum_{k=1}^n |a_k|.$$. Then $s_i-s_j=\sum_{k=1}^i |a_k|-\sum_{k=1}^j |a_k|$, not $\sum_{i=1}^n |a_i|-\sum_{j=1}^n |a_j|$. – choco_addicted Feb 22 '16 at 5:23
• One more problem: $\sum_{k=1}^n r^k$ converges to $\frac{r}{1-r}$ as $n\to\infty$, not $\frac{1-r^{n+1}}{1-r}$. – choco_addicted Feb 22 '16 at 5:29

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.
$$\sum_{k=0}^\infty a_k$$ converges absolutely, consider the sum $$|\sum_{k=0}^\infty a_k|$$ = $$|\sum_{k=0}^N a_k + \sum_{k=N}^\infty a_k|$$ $$\le$$ $$|\sum_{k=0}^N a_k| + |\sum_{k=N}^\infty a_k|$$
Notice that the first sum is convergent because it has finite terms. The second sum is where you use the assumption given about each $$\|a_k|^\frac1k$$ < r and that 0 $$\le$$ r < 1.