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.

Suppose $\sum_{i=1}^\infty \frac{1}{4i + 1}$ converges. Then as for all $i \geq 1$,
$$\frac{1}{4i + 1} > \frac{1}{4i + 2} > \frac{1}{4i + 3} > \frac{1}{4i + 4} $$
it must be the case each of these series also converge: 
$$\sum_{i=1}^\infty \frac{1}{4i + 2}, \quad \sum_{i=1}^\infty \frac{1}{4i + 3}, \quad \sum_{i=1}^\infty \frac{1}{4i + 4}$$
As everything in sight is positive and we can rearrange terms, we now have that
$$\sum_{i=1}^\infty \frac{1}{4i + 1} + \sum_{i=1}^\infty \frac{1}{4i + 2} +  \sum_{i=1}^\infty \frac{1}{4i + 3}+  \sum_{i=1}^\infty \frac{1}{4i + 4} = \sum_{n=5}^\infty \frac 1n$$
converges. But the last expression on the right does not. Contradiction.
Hence
$$\sum_{i=1}^\infty \frac{1}{4i + 1}$$
must diverge.
