Arithmetic or Geometric sequence?

Question

Given a sequence:

$$1, \frac12, \frac13, \frac14, \frac15,...$$

Its explicit formula can be given as:

$a(n) = \frac1n$ where $n \ge 1$.

I actually want to know is it a geometric sequence or an arithmetic one?

I tried finding common ratio and the common difference in this sequence to see if it's either one of them but was not successful.

My common ratio ($r$) and common difference ($d$) were some horrible values.

Answer 1

The sequence you gave is called the Harmonic sequence. It is neither geometric nor arithmetic. Not all sequences are geometric or arithmetic. For example, the Fibonacci sequence $1,1,2,3,5,8,...$ is neither.

A geometric sequence is one that has a common ratio between its elements. For example, the ratio between the first and the second term in the harmonic sequence is $\frac{\frac{1}{2}}{1}=\frac{1}{2}$. However, the ratio between the second and the third elements is $\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3}$ so the common ratio is not the same and hence this is NOT a geometric sequence.

Similarly, an arithmetic sequence is one where its elements have a common difference. In the case of the harmonic sequence, the difference between its first and second elements is $\frac{1}{2}-1=-\frac{1}{2}$. However, the difference between the second and the third elements is $\frac{1}{3}-\frac{1}{2}=-\frac{1}{6}$ so the difference is again not the same and hence the harmonic sequence is NOT an arithmetic sequence.

Answer 2

This is not a geometric series.

$a_1=1, a_2=\frac12, a_3=\frac13$

If this is a geometric sequence, then it is necessary that $\frac{a_2}{a_1}=\frac{a_3}{a_2}$

$$\frac{a_2}{a_1}=\frac{1}{2}$$

$$\frac{a_3}{a_2}=\frac{2}{3}$$

The two numbers are different, hence it is not a geometric sequence.

Similarly, you can verify that $$a_2-a_1 \neq a_3-a_2$$

To prove that something is a geometric sequence, you have to show that $\frac{a_{n+1}}{a_n}$ is a constant.
To prove that something is an arithmetic sequence, you have to show that $a_{n+1}-a_n$ is a constant.

For this problem, $$\frac{a_{n+1}}{a_n}=\frac{1/(n+1)}{1/n}=\frac{n}{n+1}=\frac{1}{1+1/n}$$ which is dependent on $n$.

while

$$a_{n+1}-a_n = \frac{1}{n+1}-\frac{1}{n}=-\frac{1}{n(n+1)}$$

which is again dependent on $n$.