# How do you show that the series 1/4+1/7+1/10+...+1/(3n+1)+... diverges?

##### 1 Answer
Feb 5, 2017

The series:

${\sum}_{n = 1}^{\infty} \frac{1}{3 n + 1}$

is divergent.

#### Explanation:

Let ${s}_{n}$ be the $n$-th partial sum of the series, and consider the sums with index $n = {3}^{k} - 1$:

${\sigma}_{k} = {s}_{{3}^{n} - 1} = {\sum}_{i = 1}^{{3}^{n} - 1} \frac{1}{3 i + 1}$

Now we evaluate the difference between two consecutive such partial sums:

${\sigma}_{k + 1} = {\sigma}_{k} + \frac{1}{3 \left({3}^{n}\right) + 1} + \frac{1}{3 \left({3}^{n} + 1\right) + 1} + \ldots + \frac{1}{3 \left({3}^{n + 1} - 1\right) + 1}$

We can see that all the terms that were added are larger than $\frac{1}{3} ^ \left(n + 2\right)$ and their number is:

${3}^{n + 1} - 1 - {3}^{n} + 1 = 2 \cdot {3}^{n}$, so we have:

${\sigma}_{k + 1} \ge {\sigma}_{k} + \left(2 \cdot {3}^{n}\right) \cdot \frac{1}{3} ^ \left(n + 2\right)$

or:

${\sigma}_{k + 1} \ge {\sigma}_{k} + \frac{2}{9}$

Now if we start from:

${\sigma}_{1} = {s}_{2} = \frac{1}{4} + \frac{1}{7} = \frac{11}{28} > \frac{2}{9}$

then we now that:

${\sigma}_{2} \ge {\sigma}_{1} + \frac{2}{9} > 2 \cdot \frac{2}{9}$

${\sigma}_{3} \ge {\sigma}_{2} + \frac{2}{9} > 3 \cdot \frac{2}{9}$

and in general:

${\sigma}_{k} > k \cdot \frac{2}{9}$

so that we have:

${\lim}_{k \to \infty} {\sigma}_{k} = \infty$

and since the ${\sigma}_{k}$ are a subset of the partial sums of the series, then the series is proven to be divergent.