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.

An elementary argument, though not the most simple or quick:  let $a_n = (1+4n)^{-1}$.  Then let $b_n = (4+4n)^{-1} = 4^{-1} (1+n)^{-1}$, so that we clearly have for all positive integer $n$, $$a_n > b_n.$$  Now note that $$S_n = \sum_{k=1}^{n-1} b_k = \frac{1}{4} \sum_{k=1}^{n-1} \frac{1}{k+1} = \frac{1}{4} \left( -1 + \sum_{k=1}^n \frac{1}{k} \right),$$ so it suffices to show that the sequence of partial sums $$H_n = \sum_{k=1}^n \frac{1}{k}$$ for $n = 1, 2, \ldots$ increases without bound.  Indeed, the familiar proof of this fact is not hard to see: $$H_{2^{m+1}} = H_{2^m} + \sum_{k=2^m+1}^{2^{m+1}} \frac{1}{k} \ge H_{2^m} + \sum_{k=2^m+1}^{2^{m+1}} \frac{1}{2^{m+1}} = H_{2^m} + \frac{1}{2}. $$  Since $H_{2^0} = 1$, it follows that $$H_{2^m} > 1 + \frac{m}{2}$$ for every positive integer $m$, consequently $$S_{2^m} > \frac{m}{8};$$ thus $S_n$ has no upper bound, and $\sum_{k=1}^{n-1} a_k$, being larger than $S_n$, must also diverge as $n \to \infty$.
