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.
 A: $$\sum_{i=1}^{n} \frac{1}{1+4i}
=\frac14\sum_{i=1}^{n} \frac{1}{i+\frac14}
\gt\frac14\sum_{i=2}^{n} \frac{1}{i}
$$
(note that the sum
goes from $2$ to $n$,
not from $1$ to $n$)
and this last sum
diverges like
$\ln(n)$,
so the original sum diverges.
A: 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.
A: 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 $n \to \infty$.
A: Try integral test for $f(x)=\frac{1}{4x+5}$.
Clearly $\int_1^\infty f(x)\,dx=\infty$, while
$$
\int_{n}^{n+1} f(x)\,dx\le f(n)=\frac{1}{4n+5},
$$
and hence
$$
\infty=\int_0^\infty f(x)\,dx=\sum_{n=0}^\infty \int_{n}^{n+1} f(x)\,dx\le \sum_{n=0}^\infty f(n)=\sum_{n=0}^\infty\frac{1}{4n+5}
$$
A: 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.
