Series

This series seems to be a telescoping series where each term is of the form $\frac{1}{(3n-2)(3n+1)}$. We can decompose each fraction into partial fractions to simplify the series.

 

Let's rewrite each term:

 

$$\frac{1}{(3n-2)(3n+1)} = \frac{A}{3n-2} + \frac{B}{3n+1}$$

 

Now, let's find the values of $A$ and $B$. We'll multiply both sides by $(3n-2)(3n+1)$:

 

$$1 = A(3n+1) + B(3n-2)$$

 

Let $n = \frac{2}{3}$, then we get:

 

$$1 = A + 0 \Rightarrow A = 1$$

 

Let $n = -1$, then we get:

 

$$1 = 0 + 3B \Rightarrow B = \frac{1}{3}$$

 

So, the decomposition becomes:

 

$$\frac{1}{(3n-2)(3n+1)} = \frac{1}{3n-2} + \frac{1}{3(3n+1)}$$

 

Now, let's rewrite the series using these decompositions:

 

$$\frac{1}{1 \times 4} + \frac{1}{4 \times 7} + \frac{1}{7 \times 10} + \dots + \frac{1}{37 \times 40}$$

 

$$= \left(\frac{1}{1} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{7}\right) + \left(\frac{1}{7} - \frac{1}{10}\right) + \dots + \left(\frac{1}{37} - \frac{1}{40}\right)$$

 

Now, observe how most of the terms cancel out:

 

$$= \frac{1}{1} - \frac{1}{40}$$

 

$$= 1 - \frac{1}{40}$$

 

$$= \frac{39}{40}$$

 

So, the sum of the series is $\frac{39}{40}$.