Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to correct an old quiz. I want to see if I have sufficiently corrected the second problem. The first I am still a bit unsure what to do.

(1) Show that the following series is divergent if $\alpha \in \mathbb{R}$ such that $|\alpha|<1$.

$$\sum_{k=1}^{\infty} \frac{1}{1+\alpha^k}$$

The above I wasn't sure what to do so I left it blank.

(2) Use the root test to decide whether or not the following series converges: $$\frac{1}{2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^4}+\frac{1}{2^5}+\frac{1}{3^6}+...$$

First note: $a_n =\begin{cases} \frac{1}{2^n} \text{ if n is odd}\\ \frac{1}{3^n} \text{ if n is even } \end{cases}$.

So $\liminf a_n = \sqrt[2n]{\frac{1}{3^n}}=\frac{1}{3}$ and $\limsup a_n = \sqrt[2n-1]{\frac{1}{2^n}}=1$ So the sequence diverges. I just want to make sure I got the values correct on this one.

share|cite|improve this question
In the second problem, the numbers are going down very fast, of course we have convergence. The calculations of the limits second line from the end are not right. – André Nicolas Dec 13 '12 at 5:30
up vote 2 down vote accepted

Note that if $\alpha \in (-1,1)$, then $1 + \alpha^k \leq 2$ so that

$$ \frac{1}{1 + \alpha^k} \geq \frac{1}{2}. $$

For the second one, since $\frac{1}{3} \leq \frac{1}{2}$, then:

$$ \frac{1}{2} + \frac{1}{3^2} + \frac{1}{2^3} + \dots \leq \sum_{k=1}^{\infty} \frac{1}{2^k} = 1 $$

If you want to use the root test: Note that

\begin{align} \limsup_{n \to \infty} \sqrt[n]{a_n} = \limsup_{n \to \infty} \sqrt[n]{\frac{1}{2^n}} = \frac{1}{2} < 1 \end{align} so the series converges absolutely.

share|cite|improve this answer
Note that question (2) asks for the root test to be used. I agree that it is simpler not to use the root test, as you have done. Further, what you've done shows the OP has made a mistake. But part of the answer should be identifying and correcting that mistake. – Pete L. Clark Dec 13 '12 at 5:26
For 1, is it because the sequence is unbounded and hence the series does not converge? – emka Dec 13 '12 at 5:28
@Pete: Thank you for the comment. I will amend my post. – JavaMan Dec 13 '12 at 5:31
@jdla: Another reason for 1) is that the terms do not have limit $0$. – André Nicolas Dec 13 '12 at 5:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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