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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.

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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

1 Answer 1

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.

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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? –  ortl Dec 13 '12 at 5:28
    
@Pete: Thank you for the comment. I will amend my post. –  JavaMan Dec 13 '12 at 5:31
2  
@jdla: Another reason for 1) is that the terms do not have limit $0$. –  André Nicolas Dec 13 '12 at 5:33

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