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.