<s>[INST] Determine if the series 1/5 + 1/8 + 1/11 + 1/14 + 1/17 ... is convergent or divergent I am trying to determine if the series $\frac{1}{5} + \frac{1}{8} + \frac{1}{11} + \frac{1}{14} + \frac{1}{17} $ is convergent or divergent?   
I can see that the terms in the denominator differ by 3 each time.  
 [/INST] The pattern for the denominator is found by first adding another term so that the series starts $\frac{1}{2} + \frac{1}{5} + \frac{1}{8} + ...$ then for the series $2 + 5 + 8 + 11 ... $ the general nth term is found as  $ T_n = a + (n-1)d$ where $a = \frac{1}{2}$ is the first term and $d = 3$ the difference. So $T_n = 3n -1$ and the original series is found to be $\sum _{n=1}^\infty \frac{1}{3n-1}$ 
If we let $f(x) = \frac{1}{3x-1}$ then $f'(x) = \frac{-1}{(3x-1)^2} $ which is negative for all $n \geq1$, which means it is decreasing on $[1,\infty)$. This function is also positive for all $n \geq1$ and so the conditions are met for the integral test. 
$$\int_1^\infty f(x) =\lim _{b\to\infty}\int_1^b \frac{1}{(3x-1)} $$
$$= \lim _{b\to\infty}\frac{1}{3}\ln(3x-1)|_1^b$$ 
$$= \lim _{b\to\infty}\frac{1}{3}[\ln(3b-1) - \ln(2)]$$
$$=\infty$$
By the integral test the series $\sum _{n=1}^\infty \frac{1}{3n-1}$ is therefore divergent.     
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