<s>[INST] Convergence of $\frac{1}{5} + \frac{1}{9} + \frac{1}{13} + ... = \sum_{i=1}^{\infty} \frac{1}{1+4i}.$ I've been working on an approximation for a problem in my numerical methods course, and I seemed to have run into the series
$$\frac{1}{5} + \frac{1}{9} + \frac{1}{13} + ... = \sum_{i=1}^{\infty} \frac{1}{1+4i}.$$
I'm trying to figure out the right test for this example. I haven't touched Calc II in a while, and I'm really not interested in getting a weird explanation from Wolfram. I was thinking that I might want to do some kind of ratio test between the $n$th and $n+1$th term.
 [/INST] Notice that:
$$(\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5})+(\frac{1}{9}+\frac{1}{9}+\frac{1}{9}+\frac{1}{9})+...\ge(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})+(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12})+..$$
So $4\sum a_n$ diverges, hence also $\sum a_n$ diverges.
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