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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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    The wiki page says that if the limit exists and is greater than $1$, then it diverges. For this particular problem, the limit does not exist. sp tjat dpes not apply. Then it says if the $\limsup$ exists and is less then $1$, it converges, and that if the $\liminf$ exists and is greater than $1$, then it diverges. It says nothing about if \limsup>1.2012-03-26

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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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    Well I could just add a little note on that (if $k$is odd, $k+1$ is even and vice versa) or distinguish between the two cases for k's parity..2012-03-26