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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.

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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.

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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

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