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.


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.

  • $\begingroup$ 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 $\sum_{i=N}^j a_i - \sum_{i=N}^k a_i$? $\endgroup$ – choco_addicted Feb 22 '16 at 5:06
  • $\begingroup$ @choco_addicted that should be $\sum_{j=N}^{n}a_j-\sum_{k=N}^{n}a_k$ ? $\endgroup$ – Simple Feb 22 '16 at 5:17
  • $\begingroup$ 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 $\sum_{i=1}^n |a_i|-\sum_{j=1}^n |a_j|$. $\endgroup$ – choco_addicted Feb 22 '16 at 5:23
  • $\begingroup$ One more problem: $\sum_{k=1}^n r^k$ converges to $\frac{r}{1-r}$ as $n\to\infty$, not $\frac{1-r^{n+1}}{1-r}$. $\endgroup$ – 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.

The proposition you have to prove is called the 'root test'.


Since you are trying to prove that

$$\sum_{k=0}^\infty a_k$$ 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.

Invoking the triangle inequality, the original assumption, the ratio test and the root test should show that you have a sum of two convergent series.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.