Convergent? - $a_{k+1} = \left( \frac{3}{4} + \frac{(-1)^k}{2} \right) a_k$ 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.

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.
