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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.
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.
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 $n$.
