# Convergence of $\sum a_n$ where $a_n$ is defined recursively

Let $a_1 = 1$ and $$\frac{a_{n+1}}{a_n} = \frac{3}{4}+\frac{\left(-1\right)^n}{2}$$

Question: How to show that $\sum a_n$ converges, i.e. $$\sum_{n=1}^{\infty} a_n< \infty$$

I am able to show, that at least $a_n\rightarrow 0$ which is a necessasy condition for the convergence:

First the subsequence $\left(a_{2k}\right)$ tends to zero because it holds

$$\frac{a_{2k+2}}{a_{2k}}=\frac{a_{2k+2}}{a_{2k+1}}\cdot \frac{a_{2k+1}}{a_{2k}} = \frac{5}{4}\cdot\frac{1}{4}=\frac{5}{16}<1$$

Now note that $$a_{2k+3} = a_{2k+2}\cdot \left(\frac{3}{4}+\frac{1}{4}\right)=a_{2k}\cdot\frac{5}{16}\cdot\frac{5}{4}=a_{2k}\frac{25}{64}$$ Hence $a_{2k+3}$ and therefore $a_{2k+1}$ and consequently $a_k$ tends to zero as well.

However, I was not able to show the convergence of the series mentioned before. I tried to apply ration test, root test but I did not found an answer. I thought about comparing $\sum a_n$ with some kind of geometric sum, but I did not found an appropriate expression to compare with

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Note that the even-numbered terms of your sequence form a geometric progression, as do the odd-numbered terms, and both progressions have ratio smaller than $1$ (as you note, $5/16$, in fact), so sum the odd terms and the even terms separately.

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Hint.

Show that the series $\sum a_{2k}$ and $\sum a_{2k + 1}$ (consisting respectively of the even- and odd-numbered terms of $\sum a_n$) are both convergent geometric series.

Show that this implies the convergence of $\sum a_n$.

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@user127001 Now what happens in a series where you alternate terms between two convergent series? – user119908 Jan 8 '14 at 18:24
Yes, I know, the sum of both series is equal to my series. But I also need that those series are absolute convergence what they of course are in this case. – user127.0.0.1 Jan 8 '14 at 18:32
You don't need absolute convergence. – user119908 Jan 8 '14 at 18:34