<s>[INST] Determining series formula Is there any procedure to follow when determining the function of a series? This seems simple but for I can't figure it out.

$$ \frac15 + \frac18 + \frac1{11} +\frac1{14} + \frac1{17}+\cdots$$

 [/INST] Look at the denominators: $5, 8, 11, 14, 17, \ldots$ Can you find an expression for these?
Start by looking at the differences: $8-5=3$, $11-8=3$, $14-11=3$ and $17-14=3$. To get from one term to the next, we simply add $3$; the term-to-term rule is $+3$. That means that the sequence $5,8,11,14,17,\ldots$ is like the three times table, i.e. it resembles $3n$. 
However, the three times table goes $3,6,9,12,15,\ldots$ Our sequence, $5,8,11,14,17,\ldots$, is always two bigger than the three times table. Hence the $n$-th term rule is $3n+2$.
Let's check: $3\times 1 + 2 = 5$, $3\times 2 + 2 = 8$, $3\times 3 + 2 = 11$, $3 \times 4 + 2=14$ and $3\times 5 + 2 = 17$.
If the sequence $5,8,11,14,17,\ldots$ is given by $3n+2$ then the sequence $\frac{1}{5},\frac{1}{8},\frac{1}{11},\frac{1}{14},\frac{1}{17},\ldots$ is given by:
$$\frac{1}{3n+2}$$
The partials sums are then given by
$$\sum_{n=1}^p \frac{1}{3n+2}$$
Sadly, as $p \to \infty$, we don't get a sensible answer. Notice that $5n \ge 3n+2$ for all $n \ge 1$ and so 
$$\frac{1}{5n} \le \frac{1}{3n+2}$$
for all $n \ge 1$. If we can show that $\sum_{n \ge 0} 1/5n$ diverges then clearly our series will diverge. Well:
$$\sum_{n \ge 1} \frac{1}{5n} = \frac{1}{5}\sum_{n \ge 1} \frac{1}{n}$$
The sum $\sum_{n \ge 1} 1/n$ is called the Harmonic Series and is well-known to diverge. Since each and every term of $\sum_{n \ge 1}1/5n$ is less than or equal to the corresponding term in the series $\sum_{n \ge 1}1/(3n+2)$, and the series $\sum_{n \ge 1}1/5n$ diverges, it follows that the series $\sum_{n \ge 1}1/(3n+2)$ also diverges.
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