# Thread: Intergral Test for Divergence

1. ## Intergral Test for Divergence

1 + 1/6 + 1/11 + 1/16 +1/21

The answer key tells me that the pattern in the denominator is 5x-4. Can someone explain how to find the pattern in this series?

2. Originally Posted by aaronb
1 + 1/6 + 1/11 + 1/16 +1/21

The answer key tells me that the pattern in the denominator is 5x-4. Can someone explain how to find the pattern in this series?
Look at the denominators 1, 6, 11, 16 and 21. The common jump is

$\displaystyle 6=1 - 5, 11-6 = 5, 16-11 = 5$ which suggest a formula like $\displaystyle a_n = 5n + b$. Then subs $\displaystyle n = 1$ into $\displaystyle a_n$ so $\displaystyle a_1 = 1$. This give $\displaystyle b = -4$. Then check to see if you get all the numbers $\displaystyle a_1=1, a_2 =6,a_3=16$ etc.

3. Hello, aaronb!

$\displaystyle 1 + \tfrac{1}{6} + \tfrac{1}{11} + \tfrac{1}{16} + \tfrac{1}{21} + \hdots$

The answer key tells me that the pattern in the denominator is $\displaystyle 5n-4$.
Can someone explain how to find the pattern in this series?
Obviously, we examine the denominators: .$\displaystyle 1,\:6,\:11,\:16,\:21,\:\hdots$

Equally obvious is the fact that they "go up by 5."

This is an Arithmetic Sequence with first term $\displaystyle a = 1$ and common difference $\displaystyle d = 5.$

You're expected to the general term of an A.S.: .$\displaystyle a_n \:=\:a + (n-1)d$

Hence, we have: .$\displaystyle a_n \:=\:1 + (n-1)5 \quad\Rightarrow\quad\boxed{ a_n \:=\:5n-4}$