Convergence of $1 +\frac{1}{5}+\frac{1}{9} +\frac{1}{13}+\dots$ This is probably incredibly simple, but we've just started the topic, and we've just gone over geometric series, p-series, and harmonic series. Its simple when the series is given explicitly in sigma-notation, but I struggle when they don't give you the form and just give you the first few numbers. The question exactly is:
Determine whether the following series converges or diverges. Give a reason for your answer.
$$1+\frac15+\frac19+\frac1{13}\dots$$
Any tips/hints/help would be much appreciated.
 A: The series you give looks like: $$\sum_{n=0}^{+\infty} \frac1{4n+1}$$
To see this, note that each of the denominators are $1$ larger than a multiple of $4$:  $5 = 4 + 1$, and $9 = 8 + 1$, and $13 = 12 + 1$.  Presumably this pattern continues.  Then we note that $4 = 4 \cdot 1$, and $8 = 4 \cdot 2$, and $12 = 4 \cdot 3$, and so on.  So the first few terms are really:
$$1 + \frac1{4 \cdot 1 + 1} + \frac1{4 \cdot 2 + 1} + \frac1{4 \cdot 3 + 1} + \cdots$$
Being able to identify this type of pattern is a skill that will come naturally with practice.  So I definitely recommend getting lots of practice!
Anyway, you can use the Limit Comparison Test with the harmonic series to get the answer to the actual question.  Let me know if you require further guidance.

EDIT:  The way you originally thought about it can also work:
\begin{align*}
  1+\frac15+\frac19+\frac1{13}+\cdots &= 1+\frac1{5-0}+\frac1{10-1}+\frac1{15-2}+\cdots\\[0.3cm]&=1+\frac1{5\cdot 1-0}+\frac1{5\cdot2-1}+\frac1{5\cdot3-2}+\cdots\\[0.3cm]&=\sum_{n=0}^{+\infty}\frac1{5\cdot n-(n-1)}\\[0.3cm]&=\sum_{n=0}^{+\infty}\frac1{4n+1}\end{align*}
A: For all $n$, we have 
$$\frac{1}{4n+\color{red}1}\geq \frac{1}{4n+\color{red}n}=\frac{1}{5}\cdot\frac{1}{n}.$$
Thus, using the Comparison Test, the series $$\frac{1}{5}+\frac{1}{9}+\frac{1}{13}+\cdots$$ is divergent since the series $\sum_{n=1}^{+\infty}(\frac{1}{5}\cdot\frac{1}{n})$ is divergent. Since adding a finite number of terms in a divergent series does not affect divergence, the given series
 $$1+\frac{1}{5}+\frac{1}{9}+\frac{1}{13}+\cdots$$ is also divergent.
A: You could also observe that 
$$1 +\frac{1}{5}+\frac{1}{9}\cdots \geq 1 + \frac{1}{5}\left (1 +\frac{1}{2}+\frac{1}{3} +\cdots\right)$$
The term inside brackets is the harmonic series which....
A: The series can be written as :
$$\sum_{n=0}^\infty\frac{1}{4n+1}$$
The integral test is the simplest in this case. The series is a lot like the harmonic series, so one might think that integral test can be used.
Compute the integral:
$$\int\frac{1}{4x+1}dx=\frac{\ln(4x+1)}{4}+C$$
Then compute the improper integral:
$$\int_1^\infty\frac{1}{4x+1}dx=\lim_{n\to\infty}\frac{\ln(4n+1)}{4}-\frac{\ln5}{4}$$
Clearly, the integral is divergent. Hence, the series is also divergent.
