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.  
 A: Note that the terms are greater than the terms of $$\frac 16+\frac 19+\frac 1{12}+\frac 1{15}+\cdots =\frac 13\left(\frac 12+\frac 13+\frac 14+\frac 15+\cdots\right)$$
When we have terms of the form $\cfrac 1{an+b}$, choose a fixed $r$ with $ar\gt b$ and consider terms of the form $\cfrac 1{a(n+r)}$
A: 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.     
A: If we include the term $\frac12$ at the beginning, then each term is larger than $\frac1{3n}$. Thus any partial sum is larger than $\frac13H_n$, where $H_n=\sum_{i=1}^n\frac 1i$ are the harmonic numbers.
The harmonic numbers are known to diverge to infinity, so the sequence $\frac 13H_n-\frac12$ must also diverge. By the comparison test, your series diverges.
