Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to show that the following series is convergent, divergent?

$\displaystyle\sum_{k=0}^\infty a_k$ where $a_1 = 1$ and $a_{k+1} = \left( \frac{3}{4} + \frac{(-1)^k}{2} \right) a_k$

It's kind of related to the geometric series, the denominator of the the k-th number is $4^k$ and the numerator grows every second step $5^k$.

I would be glad to only get hints and go from there then..


Observe that $a_{k+2} = \left( \frac{3}{4} + \frac{(-1)^{k+1}}{2} \right) a_{k+1} = \left( \frac{3}{4} + \frac{(-1)^{k+1}}{2} \right) \left( \frac{3}{4} + \frac{(-1)^{k}}{2} \right) a_{k} \\= \left( \frac{3}{4} - \frac{1}{2} \right) \left( \frac{3}{4} + \frac{1}{2} \right) a_{k} = \frac{1}{4} \cdot \frac{5}{4} a_k = \frac{5}{16} a_k $

$\displaystyle \sum_{k=0}^\infty a_k = \sum_{k=0}^\infty b_k + \sum_{k=0}^\infty c_k $

Where $b_1 = 1$ and $b_{k+1} = \frac{5}{16} b_k$ and $c_1 = \frac{1}{4}$ and $c_{k+1} = \frac{5}{16} c_k$.

The latter two are convergent according to the ratio test, because $\lim\sup \frac{|b_{k+1}|}{|b_k|} < 1$ and $\lim\sup \frac{|c_{k+1}|}{|c_k|} < 1$ therefore the original series is convergent as well.

share|improve this question
    
Just noticed, the quotient $\frac{a_{n+1}}{a_n}$ is either $\frac{1}{4}$ or $\frac{5}{4}$. So maybe use the ratio test, where $\lim\sup$ is $\frac{5}{4}$ and therefore the series is divergent? –  kannn Mar 26 '12 at 16:24
    
That's the right idea, if wrong answer. –  Thomas Andrews Mar 26 '12 at 16:25
    
Hm.. where's my error in reasoning.. –  kannn Mar 26 '12 at 16:27
    
Why can you conclude that just because one of the terms is $5/4$ that it diverges? If that were true, write out $3z + 3z^2 + 3z^3 + ...$ as $z + 2z + z^2 + 2z^2 + ...$. By your reasoing, since $2z$ is twice $z$, and $2z^2$ is twice $z^2$, etc., this can never converge. But it does when $|z|<1$ –  Thomas Andrews Mar 26 '12 at 16:30
    
I just noticed: Our phrasing of the ratio test just includes the $\lim\sup$ and say if $\lim\sup > 1$ then the series diverges. Again when I look at the $\lim\sup \frac{a_{n+1}}{a_n}$ I can't see why the $\lim\sup$ shouldn't equal $\frac{5}{4}$ and therefore is greater than 1. The phrasing in wikipedia useses $\lim$ or a more diverse phrasing with $\lim\sup$ and $\lim\inf$ - confused.. –  kannn Mar 26 '12 at 18:04

1 Answer 1

Hint: Show $a_{k+2} = \frac{5}{16} a_k$ for all $k$.

What does this say about $a_1 + a_3 + ... + a_{2n-1} + ...$?

What does it say about $a_2 + a_4 + ... + a_{2n} + ... $?

An alternate approach is to not that if $d_k=a_{2k-1}+a_{k2}$, then:

$$d_{k+1} = a_{2k+1} + a_{2k+2} = \frac{5}{16}(a_{2k-1} a_{2k}) = \frac{5}{16}d_k$$

Now, in general, just because $(a_1+a_2) + (a_3+a_4) + ...$ converges, it doesn't mean that $a_1+a_2+...$ converges. For example:

$$(1+(-1)) + (1+(-1)) + ... $$

coverges, but

$$1 + (-1) + 1 + (-1) ... $$

does not.

However, this is true of all the $a_i$ are positive, as in this case.

So the fact that $\sum_{k=1}^\infty d_k$ converges would mean that $\sum_{k=1}^\infty a_k$ converges.

share|improve this answer
    
Both convergent? According to the ratio test the former and the latter always have the ratio $\frac{5}{16} < 1$ –  kannn Mar 26 '12 at 16:29
    
Yes, so what can you say about $a_1+a_2+ ...$? –  Thomas Andrews Mar 26 '12 at 16:32
    
I tried to include this in my original question.. I can see that it's $\frac{5}{16}$ but I think I need to show this step waterproof.. I'll try to add this.. –  kannn Mar 26 '12 at 16:40
    
To show it "waterproof", just look at the two separate cases, when $k$ is odd and when $k$ is even. –  Thomas Andrews Mar 26 '12 at 16:42
    
I edited my question. I think this should be ok without paying attention to the parity of $k$ since if $k$ is odd then $k+1$ is even and vice versa. What do you think?<br>Hey, realized that this trick to calculate $a{k+2}$ to get around this alternating $(-1)^k$ could be useful for this kind of series.. –  kannn Mar 26 '12 at 16:48

Your Answer

 
discard

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.